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Part II –Thermodynamic propertiesProperties of Pure Materials – Chapter 8

Primary objective is to evaluate changes in statein terms of primitive and measurable properties

Secondary objective is to connect molecular properties and interactions to macroscopic properties and processes

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Properties of Pure Materials – Chapter 8

• Connections to the Fundamental Equation via G• various forms for equations of state• reference state condition for S = So and the

Third Law of Thermodynamics • Derived property estimation U and ∆U, etc.• Role of departure or residual functions• Constitutive PVTN volumetric property models

Ideal gas lawTheorem of Corresponding States Fluid behavior from the Boyle point to the triple point –Zeno condition Pressure and volume explicit semi-empirical EOSsCorrelated experimental data

• Ideal gas state heat capacity modelstranslation – kinetic theory - classicalrotation – rigid rotator -classicalvibration – quantized using the Einstein model

• Property estimation methods Molecular group contributionsCorresponding StatesConformal fluid theoryMolecular simulationswww.bsscommunitycollege.in www.bssnewgeneration.in www.bsslifeskillscollege.in


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Connections to the Fundamental Equation

U = TS – PV + µ N = f ( S,V, N)

dU = TdS – PdV + µ dN

G = U + PV –TS = H – TS = y(2) = f(T,P, N )

dG = -S dT +VdP + µ dN

κ

α

∂ ∂⎛ ⎞ ⎛ ⎞≡ =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂⎛ ⎞≡ − ⎜ ⎟∂⎝ ⎠∂⎛ ⎞≡ ⎜ ⎟∂⎝ ⎠

1

1

pP P

TT

PP

S HC TP P

VV P

VV T



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Constitutive PVTN Volumetric Property Models

1. Ideal gas law PV = NRT or PV = RTZ = PV/RT = 1

2. Theorem of Corresponding StatesZ = f ( Tr , Pr , Zc , ω, …)scaling to reduced coordinatesfluids from the Boyle point at low density tothe triple point at high density – the Zeno line

3. Cubic type EOS P = f(T,V)van der Waals P = RT/(V-b) –a/V2

Redlich-Kwong (RK)Redlich- Kwong-Soave (RKS)Peng-Robinson (PR)

4. Virial type EOS Z = 1 + B/V + C/V2 + …BWRStarlingMartin-Hou



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Combinatorics When working with particles on a lattice (whether it be 1, 2, or 3 dimensional), use the equations below to determine Ω, the number of possible configurations the system can take. For M = # of sites N = # of particles in system Case #1: Particles are : non-interacting (more than one can occupy same site) distiguishable

NMΩ = Case #2: Particles are : exclusionary (only one can occupy a given site) distiguishable

M !( M N )!

Ω =−

Case #3: Particles are : exclusionary (only one can occupy a given site) indistiguishable

M !N !( M N )!

Ω =−

Case #4: Particles are : non-interacting (more than one can occupy same site) indistiguishable

( )11

M N !N !( M )!

+ −Ω =

−

These equations can be used to determine the entropy for the microcanonical ensemble. For this ensemble, since each microstate has the same energy, the entropy is determined solely by the number of possible configurations of the system:

config ln S S k= = Ω



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10.40 DEFINITIONS _________________________________________________________________________

SYSTEM/SURROUNDINGS (ENVIRONMENT) STATE OF A SYSTEM - IDENTIFIED BY PROPERTY VALUES REQUIRED REPRODUCE THE SYSTEM SIMPLE SYSTEM - DEVOID OF ANY INTERNAL ADIABATIC, RIGID, IMPERMEABLE BOUNDARIES, NO EXTERNAL FORCE FIELDS OR INERTIAL FORCES, CAN BE SINGLE OR MULTI-PHASE COMPOSITE SYSTEM - 2 OR MORE SIMPLE SYSTEMS PHASE - REGION OF UNIFORM PROPERTIES EXTENSIVE/INTENSIVE PROPERTIES - FIRST/ZERO ORDER IN MASS PRIMITIVE PROPERTY - MEASURABLE DERIVED PROPERTY - DEFINED IN TERMS OF CHANGES IN THE STATE OF A SYSTEM OPEN VERSUS CLOSED SYSTEMS - WITH RESPECT TO MASS FLOW STATE VERSUS PATH FUNCTIONS ISOLATED SYSTEM - NO INTERACTIONS WITH SURROUNDINGS QUASI-STATIC, REVERSIBLE, AND IRREVERSIBLE PROCESSES BOUNDARIES

-ADIABATIC/DIATHERMAL -PERMEABLE/IMPERMEABLE/SEMI-PERMEABLE -RIGID/MOVABLE



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10.40 PRIMARY NOMENCLATURE/SYMBOLS

VARIABLE INTENSIVE EXTENSIVE ____________________________________________________________________________ GENERAL PROPERTY B B ENERGY E E HEAT - - - Q HEAT CAPACITY Cv ,Cp - - - WORK - - - W TEMPERATURE T - - - TIME - - - t PRESSURE P - - - VOLUME V V INTERNAL ENERGY U U ENTHALPY H H ENTROPY S S HELMHOLTZ FREE ENERGY A A GIBBS FREE ENERGY G G CHEMICAL POTENTIAL µ - - - MOLE FRACTION/MOLES x, y n, N FUGACITY f - - - FUGACITY COEFFICIENT φ - - - ACTIVITY a - - - ACTIVITY COEFFICIENT γ - - -



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Energy and the 1st law Treatment Summary of mathematical forms

System

Expression

1. closed, general WQEd δ+δ= 2. closed, simple WQUdEd δ+δ== 3. closed, simple,

only PdV work VPdQUd −δ=

4. open, simple outoutout

ininin

nHnHWQUd δ∑−δ∑+δ+δ= σσ

5. open, non-simple

PE+KE only (use mass basis)

outout

outoutout

inin

ininin

mvgzHmvgzHWQEd δ⎟⎟⎠

⎞⎜⎜⎝

⎛++∑−δ⎟

⎟⎠

⎞⎜⎜⎝

⎛++∑+δ+δ= σσ 22

22

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛ ⟩⟨+⟩⟨+=

2

2vmzmgUdEd

6. steady state, open, non-simple

nt

nt

ndtdN

dtdm

dtEd

dNdmEd

outin =δ

δ=

δδ

===

===

0

reactions) no if( 0or 0 0

plus expression (5)

( )nKEPEHt

Wt

QdtEd

ssssss ∆+∆+∆=δ

δ+

δδ

= σσ 0

( )2

22inout

ssinoutssinoutssvvKEzzgPEHHH −

=∆−=∆−=∆

7. non-simple, open distributed interaction along σ-surface

( ) 2

, 2iV a a

E vdV d W H gz dt

a aσ σ

σ σ σ

ρρ

∂⎛ ⎞ ⎛ ⎞= − ⋅ + ∑ − + + ⋅⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠⎝ ⎠

∫ ∫ ∫q n v n



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Four Postulates can be used to define classical thermodynamics: 1. For closed simple systems with given internal restraints, there exist stable equilibrium states that can be characterized completely by two independently variable properties in addition to the masses of the particular chemical species initially charged.

- "closed, simple systems" - "independently variable properties" - masses of each component known - consistent with the Gibbs phase rule - restatement of Duhem's theorem

L - V coexist

L

S

Crit. Pt.

L

V

Crit. Pt.

P S

S+V

L

L+V

S+L

S

S+V

L

L+V

S+L

P

T V

F = n+2-π = 3- π for pure n=1 system



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2. In processes for which there is no net effect on the environment, all systems (simple and composite) with given internal restraints will change in such a way that they approach one and only one stable equilibrium state for each simple subsystem. In the limiting condition, the entire system is said to be at equilibrium.

- applies to all isolated systems

- describes the natural tendency of processes to approach equilibrium at a minimal energy and maximal entropy state - linking postulates I and II provides a means for mathematically describing thecriteria of phase and chemical equilibria



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3. For any states (1) and (2), in which a closed system is at equilibrium, the change of state represented by (1) to (2) and/or the reverse change (2) to (1) can occur by at least one adiabatic process and the adiabatic work interaction between this system and its surroundings is determined uniquely by specifying the end states (1) and (2).

- adiabatic work interactions of any type in closed systems are state functions. - adiabatic form of the 1st law :

dE = δWadiabatic = δQ + δW where δW and δQ are the actual heat and work interactions that occur for non-adiabatic processes between the same end states (1) and(2).



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4. If the sets of systems A,B, and A,C each have no heat interaction when connected across nonadiabatic walls, there will be no heat interaction if systems B and C are also connected.

-"Zeroth law of thermodynamics"

- concept of thermal equilibrium and the absence of a heat interaction - temperature differences are needed for heat transfer to occur



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Specific type of work interaction δW = Fs • dx pressure-volume — – PdV (expansion or compression of system volume) surface deformation — σ da ( σ= surface tension, a = surface area) electric charge transfer — ∃dq ( ∃ (or ε) = electromotive force or potential, q = charge) electrical polarization — E dD ( E = electric field strength, D = electric displacement) magnetic polarization — MdB ( M = magnetization, B = magnetic induction) frictional — Ff dx stress-strain — Vo( Fx /a)dΩ (Fx /a = stress, dΩ = linear strain = dx/xo and VodΩ/a=dx) general shaft work — δWs



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Supplementary Notes for Chapters 1-3

Context and Approach 1st Law: Concepts and Applications

These notes are intended to summarize and complement the material presented in our textbook the 3rd edition of Thermodynamics and Its Applications and discussed in our graduate thermodynamics class (10.40). For the most part, we use the same notation and make references to illustrations and equations contained in the text. Hopefully you will find these notes helpful for self study and review. We strongly urge you to consider the example problems and other problems that appear at the end of each chapter from Chapter 3 onwards. We begin by summarizing the learning objectives for Chapter 1-3.

Learning Objectives* Chapter 1 After finishing this chapter you should be able to: • Describe the three general types of problems that are encountered in thermodynamics • List the basic problem solving steps and identify the types of information required to solve

thermodynamics problems • Understand the difference between the postulatory and historical approach to

teaching/learning thermodynamics Chapter 2 After finishing this chapter you should be able to: • Clearly and without jargon define: system, boundary, environment, primitive properties,

derived properties, thermometric temperature, event interaction, adiabatic, diathermal, phase, restraint, thermodynamic processes, path and state

• Explain the difference between an open, closed and isolated system • Give an example of an adiabatic wall and explain why it is useful • Describe and give examples of a simple and a composite system • Understand what is meant by a stable equilibrium state and how it may be characterized • Explain the difference between intensive and extensive properties • Describe situations where thermodynamic properties are not independently variable • Use the nomenclature and unit systems employed in engineering thermodynamics *contributed in part by Professor Brad Eldredge, University of Idaho



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Chapter 3 After finishing this chapter you should be able to: • Clearly define and give examples of work, adiabatic work interactions, heat interactions,

ideal gas, energy, internal energy, enthalpy and heat capacity • Be able to calculate PdV work interactions for quasi-static, closed system expansions and

compressions • Understand the connection between Joule’s experiments and adiabatic work interaction • Write the complete general form of the 1st Law and show how to simplify it for simple open

and closed systems and for steady state and transient situations • Using the 1st Law, calculate the change in energy for a closed systems going from State 1 to

State 2 and determine the heat and/or work required to effect this change • Using the 1st Law, calculate the change in energy for open system undergoing a change from

State 1 to State 2 and determine the heat, work and/or mass flows required to make this change

Postulatory versus Historical Approach

Postulatory – axioms that cannot be proved from first principles are stated – they represent the minimum set of rules, consistent with the experimental body of knowledge, that determine how systems behave in terms of the interchange of energy (heat, work, ... ) and mass. Historical – chronological development of concepts (and misconceptions) based on an evolving set of experimental knowledge. Postulates tend to be abstract – but with continued exposure they become the logical set of rules needed to formulate the thermodynamic laws and relationships that govern the behavior of systems that transfer energy by heat and work interactions and by mass transfer. In addition, the postulates provide a means of specifying the distribution of components in phase and chemical equilibrium as well as the stability of specific phases. Within the context of the historical development, our predecessors faced problems and situations that they couldn’t explain with existing principles in physics and mathematics. Consequently, they proceeded to create new theories and “laws” verifying their hypotheses with experiments. The amalgamation of their work is captured in the four main postulates of our text. In table I and the two figures that follow, major contributors to classical chemical and statistical thermodynamics are briefly described in a historical context.



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Major Early Contributors to Classical Chemical and Statistical Thermodynamics

Scientist

Black Calorimetry

Mechanical equivalent of heat

Heat engines and conversion of heat into work

Interconversion of heat and work

Interconversion of heat and work

Absolute temperature scale

2nd law concepts

2nd law, dissipation of energy

Concept of entropy

Heat, statistical basis for 2nd law, clarification of many key issues

Thermodynamic properties from distribution functions

Chemical thermodynamics

Chemical potential; phase rule; fundamental equation

Defined classical statistical thermodynamics; created ensembles

1798

1760-1766

1824

1842

1843-1852

1848-189

1850

1851

1854-1865

1870-1879

1871

1873

1876-1878

1902

1882

1884-1887

1890

1909

1906

1927

Count Rumford(B. Thompson)

Carnot

Mayer

Joule

W. Thomson(Kelvin)

W. Thomson(Kelvin)

Clausius

Clausius

Maxwell

Boltzmann

Gibbs

Gibbs

Gibbs

Helmholtz

van=t Hoff

Duhem

Caratheodory

Nernst

Simon

Years Contributions

Theory of equilibrium; free and bound energy ( A defined)

Chemical thermodynamics; theory of equilibrium constant; solutions

Mixtures

Adiabatic work and 1st law

Heat theorem (3rd law)

Improved version of 3rd law

Adapted from Laidler, K.J. (1993). The World of Physical Chemistry, Oxford.



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GuggenheimFowler

Gibbs

Hilderbrand PitzerWilsonRenonet al.

ScatchardRedlich-KisterPrausnitzMeissner

van derWaals

van LaarMargulesG.N. LewisRandall

DebyeHuckel

FloryHuggins

ClausiusOnnesMayer

BeattieBridgemanBWR Starling et al.

modern statistical mechanics

cubic EOS

GEX - models

electrolyte models

polymer EOS models

Virialtheorem

VirialEOS

ClusterIntegrals

Tait, Hartman, Sanchez-Lacombe, et al.

Mayer, HCB, Chandler,McQuarrie, et al.

Redlich-Kwong, Soave, Peng-Robinson,Martin, et al.

Pioneer and inventor - of classical statistical and chemical thermodynamics, analysis of multicomponent, multiphase systems, ensembles, Fundamental Equation, phase rule, chemical potential, etc.

Modern Classical Era Contributors



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Historical Progression of Classical Thermodynamics

Preclassical Era

Mostly physics - focus on experiments

Mostly physics, math, and mechanical engineering - focuson laws and postulates

Chemical and molecularthermodynamics - focus onnon-ideal fluids, phase andchemical equilibrium andstability and so forth

GalileoMaxwellClausiusLord KelvinBoltzmann

BlackCount Rumford

Joule & Carnot1600 1840 1900 2000?

Work and HeatConcepts

Statistical and Chemical Thermodynamics

Gibbs

Classical Era Modern Classical Era



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Discussion of Postulates (listed in Appendix A) 1. For closed simple systems with given internal restraints, there exist stable equilibrium states that can be characterized completely by two independently variable properties in addition to the masses of the particular chemical species initially charged.

- "closed, simple systems" - "independently variable properties" - masses of each component known - consistent with the Gibbs phase rule - restatement of Duhem's theorem

2. In processes for which there is no net effect on the environment, all systems (simple and composite) with given internal restraints will change in such a way that they approach one and only one stable equilibrium state for each simple subsystem. In the limiting condition, the entire system is said to be at equilibrium.

- applies to all isolated systems - describes the natural tendency of processes to approach equilibrium at a minimal energy and maximal entropy state - linking postulates I and II provides a means for mathematically describing the

criteria of phase and chemical equilibria

3. For any states (1) and (2), in which a closed system is at equilibrium, the change of state represented by (1) to (2) and/or the reverse change (2) to (1) can occur by at least one adiabatic process and the adiabatic work interaction between this system and its surroundings is determined uniquely by specifying the end states (1) and (2).

- adiabatic work interactions of any type in closed systems are

state functions. - adiabatic form of the 1st law : dE = δWadiabatic = δQ + δW

where δW and δQ are the actual heat and work interactions that occur for non-adiabatic processes between the same end states (1) and (2).

4. If the sets of systems A,B, and A,C each have no heat interaction when connected across nonadiabatic walls, there will be no heat interaction if systems B and C are also connected.

-"Zeroth law of thermodynamics" - concept of thermal equilibrium and the absence of a heat interaction - temperature differences are needed for heat transfer to occur

L - V coexist


P

T

L S

Crit. Pt.L

V

Crit. Pt.

P

V

S S+V

L L+V S+L S

S+V

L L+V S+L



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Work Interactions In general, all work interactions are path dependent and are defined to occur at the boundary of a system. The symbol δ is used here to designate a path dependent property. Work in general can be represented by the dot product of a boundary force Fs and an extensive displacement x which can be expressed differentially using dx. Thus,

δW= Fs• dx = Fσ boundary dx Sign convention for work: W or δW > 0 if the surroundings does work on the system. In this manner it is possible to maintain a consistent convention that work transferred into a system will increase its energy content and work transferred out to the surroundings will decrease its energy content Specific type of work interaction δW = Fs • dx pressure-volume — –PdV (expansion or compression of system volume) surface deformation — σ da ( σ= surface tension, a = surface area) electric charge transfer — ∃dq ( ∃ (or ε) = electromotive force or potential, q = charge) electrical polarization — E dD ( E = electric field strength, D = electric displacement) magnetic polarization — MdB ( M = magnetization, B = magnetic induction) frictional — Ff dx stress-strain — Vo( Fx /a)dΩ (Fx /a = stress, dΩ = linear strain = dx/xo) and VodS/a=dx general shaft work — δWs Pressure-volume work: Consider a piston cylinder geometry with quasi-static conditions, thus the path can be described differentially. Define the gas inside the cylinder below the piston as the system, in this case the work effect at the piston-gas boundary is, δWsystem = δWg = ! Pgas dV gas = Fg dx = ! Pg a dx We can also describe δWg using external effects by employing a force balance to connect Fg to effects in the surroundings:



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gas

atmos.

Piston of area a

Fa = Paa

mv dv/dzFg

mg Ff

z

fag FdzdvmvmgPF a +++=

dzFmvdvmgdzdzPxdFW faggas a +++=⋅=δ where dzFWmvdvmgdzWdzPW fwallpistonaatmos a =+=δ= d and ; .d , i.e. frictional dissipation at the wall. Integrating,

dzFvvmzzmgPW fagas a ∫+−+−+= )(2

))(( 21

2212

Note that dzF f may be small and can safely be neglected in many cases. Alternative conceptualization--a piston-cylinder arrangement with the gas contained selected as the system. Small weights are removed and placed in storage in a gravitational field in such a manner that the P-V history of the gas expansion is reproduced exactly. This effectively is a quasi-static process where the net work is equal to the ∆ΡΕ contained in the stored weights. Thus we can always equate the work done by a system to the rise or fall of weights (differentially) in the surroundings In other words, 0dd =∑+∑ systemgssurroundin WW as a direct result of the dynamic force balance that maintains 0=+ ∑∑ gssurroundinsystem FF

massless piston

gas z



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Adiabatic Work Interactions and Postulate III Having developed the definitions of various types of work interactions, we can now examine the consequences of Postulate III in detail. First, we need to establish a convention for the sign of a work interaction. In 10.40 we assume that work is positive if the surroundings do work on the system to increase its energy content.

δW > 0 and with Postulate III dE = + δWadiabatic and ∆E = + Wadiabatic Now what exactly is an adiabatic work interaction? Here we must rely on some rather specific definitions. We recommend that you carefully read section 3.2 to firm up your understanding of adiabatic work interactions. Then proceed to examine the third postulate in the context of a change in state from (1) to (2) (or from (2) to (1)) for a closed system given by a change in the total system energy (E) and represented by an adiabatic process occurring at least in one direction from (1) to (2) or from (2) to (1). Historical Perspective -- Joule's Experiments The 1st Law of thermodynamics resulted in part from a series of experiments carried out by Joule between 1843 and 1848. His experiments dealt with adiabatic work interactions and showed an almost exact proportionality between the amount of work expended on a system and the rise in temperature observed in a fixed mass of liquid water. Joule used a paddle wheel (shaft work) in one experiment, the compression of a gas (PdV work) in another, an electric current (electrical work) in a third test, and frictional work (Ff dx) in a fourth set of experiments. He viewed that the change of state of the water resulted from a conversion of work into "heat" and demonstrated that a constant conversion factor existed regardless of the type of work expended, namely that 4.184 J = 1 calorie. Since heat is only transferred across a system boundary, it is incorrect to regard the temperature increase of the water as induced by a heat interaction. Strictly speaking, it corresponds to a change in internal energy for an adiabatic system. Heat Interactions Heat in a classical sense in thermodynamics is "devoid of any microscopic (physical or molecular) significance" and is defined in terms of the difference between the adiabatic work interaction and the actual work interaction that occurs as the system changes from state 1 to state 2.

Q = Wadiabatic – W = E2 – E1 – W or δQ = dE – δW

where the energy difference E2 - E1 = ∆E is a state function which can be computed from an adiabatic work interaction. Like δW, δQ is a path–dependent function that occurs only at the boundary of a system.



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Path versus State Functions

In the above treatment two different differential operators d and δ have been used to indicate exact and path–dependent functional behavior, respectively. δ refers to the path dependent transfer of a differential amount of work or heat (and mass as well). d refers to a differential change in a state function such as E or U. When these functional differentials are integrated important differences appear. For example,

IδW = W and IδQ = Q (total amount of work or heat as a path dependent line integral)

I dE = ∆E = E2 – E1 (total change of the system energy E from state 1 to state 2 as indicated by the difference operator ∆)

I dE = 0 (total change of E around a closed cycle is zero)

I 0≠Wδ or I 0≠Qδ (in general, these line integrals are not necessarily zero)



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First Law for Thermodynamics for Simple, Open Systems entering streams — in leaving streams — out state 1 — initial condition state 2 — final condition Although the system A bounded by the σ–surface is open, the composite system A + δnin is closed. If E is the total energy of system A, then by applying the 1st law to the closed composite system:

E2 − (E1 + Ein δnin ) = δQ + δW + PinVin δ nin

If the composite system is simple then E = U in system A and Ein = Uin for the incoming stream then we can define a new property the enthalpy H of the incoming stream as

Hin / (U + PV )in

Therefore, the open system form of the 1st law can be written in differential form for this particular simple system A as:

dU = δQσ + δWσ + Hin δnin

Generalizing for multiple entering and exiting streams:

outoutin

ininin

nHHWQUd δ∑−δ∑+δ+δ= σσ (3-60)

Note that we have retained the path dependent differential operator δ for nin and nout because, strictly speaking, only dN of the system is a state variable with:

dN = δnin − δnout (3-58)

or for the case of multiple streams out

outin nndN δ∑−δ∑=

in

Image by MIT OCW.

σ surface

σ surface opento mass at thispoint

δnin

at Pin' Vin with Ein

insulation

Qσ

Wσ

Generalized Open Simple System A

Adapted from Tester, J. W. and Modell, Michael. Thermodynamics and Its Applications. Upper Saddle River, NJ: Prentice Hall PTR, 1997, p. 46.



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First Law for Open, Non-simple Systems with PE + KE Effects Decomposition of E into three main components

1. KE due to inertial effects (velocity/acceleration of center of mass) 2. PE due to location in gravitational field of the earth 3. Internal (U) – everything else associated with microscopic energy storage on a

molecular level. This uncoupling requires that we use constitutive equations for kinetic and potential energy which are easily obtained from our experience in physics and mechanics, namely:

UEEE PEKE ++= (3-61) UdEdE dEd PEKE ++= (3-62) where

2 2

and 2 2KE PE

m mvE E mgz

< >= = =

v

here m is the mass of the system or subsystem of interest and *<v>* is the magnitude of the velocity vector <v>.

By following the decomposition of E outlined above and the mathematical approach that led to Eq. (3-60), we obtain a general expression for the First Law for an open system with its center of mass located at a vertical distance <z> from a reference plane at z = 0 and traveling with velocity <v> as shown in Figure 3.11. Note that now we will employ a mass (rather than mole) basis for the intensive enthalpy terms in the summations of Eq. (3-60): that is Hin and Hout have SI units of J/kg. The incoming and outgoing streams also need their appropriate potential energy contributions (from gzin*nin/2 and gzout*nout) and kinetic energy contributions (from v2

in*nin/2 and v2

out*nout/2). For this case, the general form of the First Law for open, non-simple systems is written as:

outout

outoutout

inin

ininin

nv

gzHnv

gzHWQEd δ

++∑−δ

++∑+δ+δ= σσ 22

22 (3-63)

with 2

2m

d E d U mg z ⟨ ⟩

= + +

v

For any transient process we can use Eq. (3-58) to obtain the rate of change of mass or moles (dN/dt) for the system and differentiate Eq. (3-63) to obtain the rate of change of the system’s total energy (dE/dt). Thus,

2

2in in

in inin

Q W nd E vH gz

dt t t tσ σδ δ δ

δ δ δ

= + + + +

∑t

nvgzH outout

outoutout δ

δ

++− ∑

2

2

(3-64)



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to avoid ambiguities, a mass basis should be selected if kinetic and potential energy effects are to be included. The sums run over all incoming and outgoing streams. For operation at steady state with single inlet and outlet streams, Eq. (3-64) is greatly simplified because both

0 and 0 ==dtdN

dtEd

As for the system, the total energy E and mass/moles N are constant.

which implies that outin nn δ=δ and

ntn

tn

tn outin &=

δδ≡

δδ=

δδ

(3-66)

Therefore, Eq. (3-64) is modified for steady state conditions to:

( )nKEPEHWQt

Wt

Qssssss &&& ∆+∆+∆=+=

δδ+

δδ

σσσσ (3-67)

where

inoutss HHH −≡∆ ; ( )inoutss zzgPE −≡∆ ; 2

22inout

ssvvKE −≡∆

represent the time-invariant changes between the outlet and inlet stream quantities. In some practical situations, other forms of macroscopic energy storage besides kinetic or gravitational potential energy may be important. To account for these effects, which could include storage in rotational modes, electric or magnetic fields, and elastic deformation by stress-strain effects, it is relatively straightforward to modify Eqs. (3-63) and (3-64) (see Chapter 18). Further, you may encounter problems where incoming and outgoing stream fluxes are distributed on the σ-boundary that encloses the system. In addition, the heat flux may also be distributed on the σ-surface rather than occurring at discrete locations and there may be multiple work interactions. In these cases, an integral form of Eq. (3-64) should be used:

Image by MIT OCW.

σ Surface

σ Surface Opento Mass at thisPoint

δnin

δnoutat Pout' Vout with Eout

Insulation

Center of Mass

δQσ

δWσ

< z >

<v>

zout

zin

z = 0

Generalized open, non-simple system moving at velocity <v> in a gravitational field reference to z = 0.

Adapted from Tester, J. W. and Modell, Michael. Thermodynamics and Its Applications. Upper Saddle River, NJ: Prentice Hall PTR, 1997, p. 49.



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14

( ) 2

?2ia ai

V

E vdV d W H gz d

ta a

σ σσ σ σ

ρ⋅

∂ = − ⋅ + ∑ − + + ⋅ ∂

∫ ∫ ∫q n v n & (3-68)

where q and v represent vectors for the heat flux and fluid velocity, and n is the unit normal vector, perpendicular to the σ-surface pointing outward. Note that there are two types of integrals involved, one for the total energy over the volume V of the system and another for the heat and mass flux contributions over the surface area of the σ-boudary aF. Note that we have inserted an underbar with the term aF to emphasize that as used here it is the total extensive area. The parameters q, ( )2/2vgzH ++ρ and v are spatially dependent.



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15

Energy and the 1st law Treatment Summary of mathematical forms

System

Expression

1. closed, general WQEd δ+δ= 2. closed, simple WQUdEd δ+δ== 3. closed, simple,

only PdV work VPdQUd −δ=

4. open, simple outoutout

ininin

nHnHWQUd δ∑−δ∑+δ+δ= σσ

5. open, non-simple PE+KE only (use mass basis)

outout

outoutout

inin

ininin

mv

gzHmv

gzHWQEd δ

++∑−δ

++∑+δ+δ= σσ 22

22

( )

⟩⟨+⟩⟨+=2

2vmzmgUdEd

6. steady state, open, non-simple

nt

nt

ndtdN

dtdm

dtEd

dNdmEd

outin &=δ

δ=δ

δ

===

===

0

reactions) no if( 0or 0 0

plus expression (5)

( )nKEPEHt

Wt

QdtEd

ssssss &∆+∆+∆=δ

δ+δ

δ= σσ 0

( )2

22inout

ssinoutssinoutssvv

KEzzgPEHHH−=∆−=∆−=∆

7. non-simple, open distributed interaction along σ-surface

( ) 2

, 2iV a a

E vdV d W H gz d

ta a

σ σ

σ σ σρ

ρ∂

= − ⋅ + ∑ − + + ⋅ ∂ ∫ ∫ ∫q n v n &

11

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Supplemental Notes for Chapter 4 Second Law: Concepts and Applications

Real, Irreversible, Quasi-static, and Reversible

Real (Irreversible)

Quasi-static

Reversible

Partially quasi-static

Internally reversible

Quasi-static processes

- Along a quasi-static path all intermediate states are equilibrium states; thus from postulate I quasi-static paths for closed, simple systems can be described by two independent properties.

- From postulate II, if a system progressing along a quasi-static path is “isolated” from its environment, then the values of all properties will remain constant and equal to those just before the isolation.

- Quasi-static processes occur at finite rates but not so rapidly that the system is able to adjust on a molecular level. There would not be, in general, gradients of any intensive properties, such as temperature, pressure, density, etc.

- Expanding a gas contained in a frictionless piston (mass m, area a)-cylinder is not a quasi-static process:

Pgas

1. pull stopsPa 2. rapid expansion during which a definite dP/dz

gradient exists in the gas phase 3. piston moves rapidly as Pgas is greater than

Pa + mg a

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Without any friction present, the gas expansion will clearly not be quasi-static. If friction is present so that the expansion process occurs very slowly, dP/dz would be negligible and the properties of the gas would remain constant if the expansion process were stopped – that is, the system would stay in some stable equilibrium state. Thus, with friction present in this manner, the expansion process is quasi-static. A similar situation is encountered in the gas cylinder blowdown. The valve controls the blowdown rate, resulting in a quasi-static process for the gas contained in the cylinder.

In fact, the adiabatic tank blowdown process could be modeled as a closed system gas expansion against a massless piston that is frictionally damped to keep Pgas = Poutside. In this case, the process is quasi-static and:

(Ud = − VPd = NV Pd ) since N = constant

Ud = dT NC − = NPdVv

for an ideal gas with P = RT/V, by eliminating V, we get

dT/T = – R/Cv (dV/V) = –R/Cv ( dT/T – dP/P)

(R/Cv+1)dT/T = (R/Cv)dP/P

If we had eliminated T, then dP/P =-(1+R/Cv) dV/V and the same equations result by treating the system as open. Upon integration, we obtain the familiar relationships for a reversible, adiabatic expansion (or compression) of an ideal gas, namely,

(Κ -1)/ ΚPVκ = constant or equivalently T/Ti = (P/Pi)R/Cp = (P/Pi)

where ≡ κ C p / C and C p = Cv + Rv

Reversible Processes

- Via Postulate II, if any (real or ideal) system in a non-equilibrium state is isolated, it will tend toward a state of equilibrium.

- All real or natural processes are not reversible. Hence reversible processes are only idealizations that are very useful in showing limiting behavior. The performance of real processes is frequently compared with ideal performance under reversible conditions.

- "In a reversible process, all systems must be in states of equilibrium at all times, that is all subsystems must traverse quasi-static paths."

- A system undergoing a reversible process is no more than differentially removed from an equilibrium state – the system passes through a set of equilibrium states.

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- "A process will be called reversible if a second process could be performed in at least one way so that the system and all elements of its environment can be restored to their respective initial states, except for differential changes of second order." For example, in a reversible expansion or compression δ(δW) ≈ dPdV

- If a cyclic process A → B → A is reversible, then when the process is carried out, no changes will occur in any other bodies. For example, if A → B involves the absorption of a quantity of heat Q, then B → A will reject the same quantity Q to the environment.

- Any reversible process is also quasi-static, but the reverse is not necessarily true.

- Simple systems undergoing reversible processes have no internal gradients of temperature or pressure.

- Friction and other dissipative forces are not present in reversible processes. A truly reversible process will always require an infinitesimal driving force to ensure that energy transfer occurs without degradation, hence its rate would be infinitely slow. Therefore, a reversible process always can be shown to require a minimum a mount of work or will yield a maximum amount of work.

- Heat engines in reversible processes operate at maximum efficiency

Summary of the 2nd Law

- The 1st law involves primarily the principle of energy conservation and is not sufficient to describe how a natural process will proceed.

- The 2nd law is concerned with describing the direction is which a process can take place. For example, the flow of heat from a hot to a cold body.

- The 2nd law describes, in mathematical terms, the physical impossibility of reversing Joule's experiments. It is not possible to convert heat into an equivalent amount of work – some amount of heat must be transferred to a second body or the environment in the process of converting heat into work.

TH- Carnot heat engines operate cyclically and reversibly between two isothermal reservoirs at TH and TC

δQH δWcand a work reservoir. The efficiency of the δQCCarnot process for converting heat into work is

TC

ηc = – δWc/δQH = (TH−TC)/TH TH>TC so 0 < ηc 1.0

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− Given a reversible process where temperature changes, it is always possible to find a reversible zig-zag path consisting of adiabatic-isothermal-adiabatic steps such that the heat interaction in the isothermal step is equal to the heat interaction of the original process.

− Definition of entropy S as an derived state function

dS δ ≡ Q / T (an exact differential) rev

∆ S = ∫ Sd and Sd = 0∫

− Clausius inequality for describing heat interactions between two isothermal reservoirs at TA and TB.

δ QA / TA δ + QB / TB ≥ 0

− For any fully reversible process, the equality applies, for all others the inequality applies

− A reversible process adiabatic process occurs at constant entropy.

− For any real process;

(1) dS > 0 (system + surroundings) (2) ∆ S universe ∆ = S system ∆ + S gssurroundin > 0

− Entropy is a measure of the degradation of work producing potential.

− All natural processes in isolated, closed systems always occur in a direction that increases entropy.

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Combined First and Second Laws

1. Closed, single phase, simple systems

From the 1st Law: dE = dU = δ Q + δ W

For an internally reversible, quasi-static process with only PdV work:

δQ = δQ = STd and δW = δWrev − = VPdrev

Therefore, dU= TdS− PdV which also provides a way to determine entropy changes:

Basis: 1 mole of ideal gas of constant Cp

dS=dU/T + P/TdV

Cv PdS = dT + dVT T

with PV=RT P R dP dS = Cv dT + T P

dT − RTT P2

combining terms:

Cv + R dPdS = dT − RT P

or

1

2

1

2 2

1 P P

T TS p −=∫ lnR ln C S d ∆ = For ideal gas only

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2. Open, single phase, simple systems

For an internally reversible, quasi-static process with one component entering and leaving the system, all intensive properties must remain the same. Hence,

Tin = Tout = T Pin = Pout = P Uin = Uout = U Sin = Sout = S Vin = Vout = V

nin

n

l i

Q

W

( )

ll i l. i li i . l i i ll

Surface

out

Insu at on

rev

rev

E, V, S, T, P

Adapted from Tester, J. W. and Mode , M chae Thermodynam cs and Its App cat onsUpper Sadd e R ver, NJ: Prent ce Ha PTR, 1997, p. 87.

Image by MIT OCW.

From a mole balance, dN = δnin − δnout . Now the 1st Law, which describes an energy balance, can be written with only PdV work as:

)Ed = Ud = δQ − VPd + ( U + dN PV rev

and likewise an entropy balance can be formulated as:

Sd = δQrev / T + SdN δ + S gen δ = Q / T + SdNrev

since δS gen = 0 for this reversible case. Now by substituting δ Qrev into the 1st Law expression:

)Ud = STd − VPd + ( U + PV − dN TS = STd − VPd µ + dN

where μ is the chemical potential defined as:

μ ≡ G ≡ U + PV − TS = H − TS

This result can be generalized for a multicomponent, single phase system that is traversing a quasi-static path as,

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nUd = Td S − PdV µ ∑ + idNi

i

( ,where U is a continuous function of n +2 variables: U = VSf , Ni ) i =1,…n and

n( (Ud = (∂ U / δ S ) V , N S d ∂ + U / ∂ V ) V d ∂ ∑ + U / ∂ Ni ) V S , N j (i )

dN , i, NS i

T -P i Which is often referred to as the Fundamental Equation of Thermodynamics.

3. Availability (maximum and minimum work concepts)

As described in Section 14.1, consider a process shown at the right that interacts with a work reservoir and rejects heat to the surroundings. Other constraints are:

- reversible/quasi-static operation - steady state δ ≡ δ nin δ = noutn

thus δ N = 0 for primary system- Carnot heat engine operates cyclicly so δ Qs is at the system temperature T

i

ll iirQs

QR

Ws

nin

n

Wc

ir To

ll i l. i li il i i ll

Pr mary system

Secondary system

Sma carnot eng ne Work reservo

out

Heat reservoat

Adapted from Tester, J. W. and Mode , M chae Thermodynam cs and Its App cat ons. Upper Sadd e R ver, NJ: Prent ce Ha PTR, 1997, p. 588.

- all heat δQR is rejected isothermally at T # Image by MIT OCW.

A differential steady state 1st Law balance around the process (primary system) yields:

Ed = 0 = δQ + δWs + ( H − Hout ) δn (1)s in

Where δ Ws represents the shaft work contribution. A steady state 1st law balance around the Carnot heat engine gives:

Ed = 0 = − δQ + δQR + δWcs

Where the – minus sign on δ Qs reflects its directional change relative to the primary system, i.e., the engine is receiving heat from the system.

δQ = δQR + δWc (2)s

Where δWc represents the Carnot heat engine work contribution. A steady state entropy balance for the composite secondary system – the process and the heat engine yields:

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i

Sd = 0 + = δQR / T + ( S − Sout ) δn (3)o in

Combining equations (1), (2), and (3) to eliminate δ Q and δ QR and solving for the total s work interaction (Carnot + shaft work) gives with rearrangement:

∑δ W = δ Wtotal = ( δW + δWs ) = [( H − Hin )− T ( S − Sin )] δn (4)c out o out

Since we are dealing with a reversible process, Eq. (4) gives the maximum work per mole that could be produced (or the minimum work required). This is called the availability or exergy:

B ≡ H − S T o

∆B ≡ Hout − Hin − To ( S − Sin ) ∆ = H − To ∆ S (5)out

Now Eq. (4) can be rewritten

δδWmax = ( δW + δWs ) ∆ = n B (6)c

or by taking the time derivative to give maximum power:

∆ B ∆ n B (7)Wmax = ( ) δn / δt = ( )

Clearly, the availability is a state function in the strictest mathematical sense so the maximum (or minimum) work associated with any steady state process is also independent of the path.

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Supplementary Notes for Chapter 5 The Calculus of Thermodynamics

Objectives of Chapter 5

1. to understand the framework of the Fundamental Equation – including the geometric and mathematical relationships among derived properties (U, S, H, A, and G)

2. to describe methods of derivative manipulation that are useful for computing changes in derived property values using measurable, experimentally accessible properties like T, P, V, Ni, xi, and ρ .

3. to introduce the use of Legendre Transformations as a way of alternating the Fundamental Equation without losing information content

Starting with the combined 1st and 2nd Laws and Euler’s theorem we can generate the Fundamental Equation: Recall for the combined 1st and 2nd Laws:

• Reversible, quasi-static • Only PdV work • Simple, open system (no KE, PE effects) • For an n component system

( ) i

n

ii dNTSHVPdSTdUd ∑

=−+−=

1

i

n

iidNVPdSTdUd ∑

=µ+−=

1

and Euler’s Theorem:

• Applies to all smoothly-varying homogeneous functions f,

f(a,b,…, x,y, … )

where a,b, … intensive variables are homogenous to zero order in mass and x,y, extensive variables are homogeneous to the 1st degree in mass or moles (N).

• df is an exact differential (not path dependent) and can be integrated directly

if Y = ky and X = kx then

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f(a,b, …, X,Y, …) = k f(a,b, …, x,y, …) and

( ) ( ),...,,...,1...,..,..,,,..,...,,

yxbafyfy

xfx

xbayba=+⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

Fundamental Equation:

• Can be obtained via Euler integration of combined 1st and 2nd Laws

• Expressed in Energy (U) or Entropy (S) representation

[ ] ∑=µ+−==

n

iiinu NVPSTNNNVSfU

121 ,...,,,,

or

[ ] i

n

i

ins N

TV

TP

TUNNNVUfS ∑

=

µ−+==

121 ,...,,,,

The following section summarizes a number of useful techniques for manipulating thermodynamic derivative relationships Consider a general function of n + 2 variables

( )3 2nF x, y,z ,...,z +

where x ≡ z1, y ≡ z2. Then expanding via the rules of multivariable calculus:

2

1

n

ii i

FdF dzz

+

=

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠∑

Now consider a process occurring at constant F with z3, .., zn+2 all held constant. Then

3 3

0y ,z ,... x ,z ,...

F FdF dx dyx y

⎛ ⎞∂ ∂⎛ ⎞= = + ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

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Rearranging, we get: Triple product “x-y-z-(-1) rule” for F(x,y): ( ) ( ) ( ) 1/// −=∂∂∂∂∂∂ xFy FyyxxF example: ( ) ( ) ( ) 1/// −=∂∂∂∂∂∂ THP HPPTTH Add another variable to F(x,y):

( ) ( )( )x

xx y

FyF

φ∂∂φ∂∂

=∂∂//

/

example: ( )( ) T

CTC

THTS

HSTHPSyxF

p

p

P

P

P/1

///

thenand),(),( ==∂∂∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

=φ=

Derivative inversion for F(x,y): ( ) ( )xx FyyF ∂∂=∂∂ //1/

example: ( ) ( ) pPP CTTSST ///1/ =∂∂=∂∂

Maxwell’s reciprocity theorem: Applies to all homogeneous functions, e.g. F(x,y, ..)

yxxyy

x

x

y FForxyF

dyxF

=⎥⎦

⎥⎢⎣

⎢∂

∂∂∂=⎥

⎦

⎥⎢⎣

⎢ ∂∂∂

,..

,...

,..

,... )/()/(

example:

i

n

iidNVPdSTdUd ∑

=µ+−=

1

VSSVNVSVVSNS UUSPUUVT ==∂∂−===∂∂ ,, )/()/(

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Legendre Transforms:

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

σ

µ−

ξ

),(),(),(),(

),(),(

aii

ii

ii

FxN

PVTS

x

Conjugate coordinates

(extensive, intensive)

General relationship Examples

],...,,[)(],...,[ 11

)0(nm NNVSfUfunctionbasisxxfy ==

STUAytransformkxyy thi

k

ii

k −==ξ−= ∑=

)1(

1

)0()( )(

or by changing variable order to U = f (V, S, N1,…,Nn),

VPUHy +==)1(

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General relationship Examples

∑∑+==ξ+ξ−=

m

kiii

k

iii

k dxdxdy11

)( i

n

iidNVPddTSAddy ∑

=µ+−−=≡

1

)1(

or

i

n

iidNdPVSTdHddy ∑

=µ++=≡

1

)1(

∑=

=ξ−=m

iii

m xyy1

)0()( 0 )2 ( 0)2( +==+ nmwithtransformtotaly n

∑=

=ξ−=m

iii

m dxdy1

)( 0 01

)2( =µ−+−= ∑=

+n

iii

n dNdPVdTSdy

(Gibbs-Duhem Equation)

Relationships among Partial Derivatives of Legendre Transforms

ji

kk

jik

ij xxyyy∂∂

∂==

)(2)()( (Maxwell relation)

][

)0()0(

ixiii

j

xyy ⎟

⎟⎠

⎞⎜⎜⎝

⎛

∂∂

=≡ξ

ii xx

yy∂∂

∂=

1

)0(2)0(

1 )0(

21

2)0(

11 xyy

∂

∂=

]1for wellas:NB[

1

1

)0(

)1(

>=ξ

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

>ξ

=−=

iy

i

ixy

ii

i

ii

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Reordering and Use of Tables 5.3-5.5 Table 5.3 – 2nd & 3rd order derivatives of [ ] in terms of )1()1( and ijkij yy etc,)0(

iiy

Table 5.4 – Relations between 2nd order derivatives of jth Legendre transform

)0()( functionbasistheand ikj

ik yy Table 5.5 – Relationships among 2nd order derivatives of jth Legendre transform to

(j-q) transform

)( jiky

)( qjiky −

12

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Supplementary Notes for Chapter 9 Mixture Thermodynamics

Key points

Nine major topics of Chapter 9 are reviewed below:

1. Notation and operational equations for mixtures 2. PVTN EOSs for mixtures 3. General effects of mixing on heat and work interactions and state property changes 4. Gibbs-Duhem relationship and thermodynamic consistency 5. Mixing functions 6. Ideal gas mixtures and ideal solutions 7. Fugacity and activity concepts 8. Fugacity coefficients from PVTN EOS property relationships 9. Activity coefficients from ∆GEX property relationships 10. Ideal reversible work effects in mixing or separating components

1. Notation and operational equations for mixtures Partial molar properties are extensive properties that determine how derived properties change as a function of mole number or composition. For a general property, B, which could be U, H, S, V, or A, the partial molar B is defined as:

[ ], ,

⎡ ⎤∂= ⎢ ⎥∂⎣ ⎦

j

ii T P N i

BBN

(9-8)

There are a few important relationships in Chapter 9 that allow you to calculate partial molar properties given the extensive form B of the property or the intensive form B.

An important one of these is:

[ ]1 , , ,

i jj j T P x j i

BB B xx≠

⎛ ⎞∂= − ⎜ ⎟⎜ ⎟∂⎝ ⎠

∑ (9-53)

Note that for a binary, two component system that Eq.(9-53) quickly yields the tangent-intercept method of evaluating partial molar properties from a graph of B versus xj at constant T and P. For example, see Figure 9.1 and the discussion in the text.

2. PVTN EOSs for mixtures Volume and pressure explicit equations of state are commonly used to represent the volumetric properties of fluid mixtures of both gases and liquids. The key feature that distinguishes a mixture EOS from its pure component counterpart is the presence of compositional dependence. This dependence expresses itself in the form of so-called “mixing rules” that incorporate pure component EOS parameters and weight them

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proportionally to the concentration of each component following a specific mathematical recipe. With the exception of the virial equation of state no rigorous theoretical approach exists to specify a mixing rule recipe. The most common approach is to use some variation of the Lorentz-Bertlelot rules used by van der Waals and others over a century ago (see Eqs (9-17 and 9-18). In mixtures that exhibit considerable non-ideality, a binary or higher order interaction parameter δij is introduced to capture specific interactions between molecules of type i and j. Each mixture PVTNi EOS will have some prescribed recipe for its mixing rules. For example see the conventions followed for the RK and PR EOS on pages 324- 326. In situations where high level non-ideal effects are present often more complex forms for mixing rules are introduced. The Wong-Sandler and Chueh-Prausnitz rules discussed on pages 327-328 are examples of this type of mixing rule. 3. General effects of mixing on heat and work interactions and state property changes There are a few typical classes of problems that you should be able to solve. These include:

• Calculating the change in enthalpy as a result of mixing to determine a heat interaction for maintaining a constant temperature or to follow some prescribed recipe for temperature, for example, see Problem 9.2.

• Calculating a change in volume as a result of mixing to determine a change in density or the magnitude of a work interaction at constant pressure.

4. Gibbs-Duhem relationship and thermodynamic consistency For mixture data and correlations – for any property B, the partial molar quantities are interrelated through the use of the general Gibbs-Duhem relationship:

1 , ,i i

n

i ii P x T x

B Bx dB dT dPT P=

∂ ∂⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥∂ ∂⎣ ⎦ ⎣ ⎦∑

Note that B is a function of T, P, and xi for i = 1, … , n. The Gibbs-Duhem relationship can be applied to any partial molar property, such as:

i i i i i ln ln ln ln .i i iˆˆH , V , , , f , G , a , , etcµ φ γ

For a binary system, the Gibbs-Duhem relationship is frequently used to check thermodynamic consistency of thermodynamic data; for example, activity coefficients. Additionally, for a binary mixture, if you have a measurement of the partial molar property of one component as a function of composition you can determine the property for the other component.

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5. Mixing functions

( ) ( )1

1 1n

mix i i i ii

B B T ,P,x , i ,...,n x B T ,P ,x , i ,...,n+ + + +

=

∆ ≡ = − =∑ (9-68)

which is in intensive form. A similar expression results for ∆Bmix in extensive form with xi replaced by Ni. You should be familiar with the concept of a reference state (+). Strictly speaking, reference states are arbitrary, but in practice several common forms appear for xi

+ ,

T+, and P+, for example:

(1) pure i at T and P of the mixture (2) xi

+ → 0 at T and P of the mixture— that is in an infinite dilution state where Henry’s law behavior is followed

(3) a fixed composition of 1 molal or 1 molar concentration at T and P of the mixture that behaves in some ideal manner – commonly used for electrolytes (see Chapter 12)

6. Ideal gas mixtures and ideal solutions In this section of the text, a set of definitions were used to characterize non-ideal solutions in terms of a deviation from ideal behavior.

• ideal gas mixture: ( ) ln igi i i iG RT y P= µ = + λ T

• ideal solution: ( ) ln IDi i i iG RT x T= µ = + Λ ,P

where λi and Λi are constants specific to component i. The key points to remember are that partial pressure yiP is the ideal gas mixture compositionally dependent variable while for ideal condensed phase solution it is the mole fraction xi . 7. Fugacity and activity concepts For a real mixture or solution we introduce the following models:

real fluid mixture: ln i i iˆG RT f i= µ = + λ

real solution: ( ) ln i i i i iG RT a T ,P ,x+ + + += µ = + µ which lead to the following definitions of the fugacity coefficient and activity coefficient

iφ and γi , respectively:

ii

i

fˆy P

φ ≡ ---- represents a deviation from ideal gas/fluid mixture behavior

and

ii

i i i

ˆ ˆif f

ˆa f x+γ ≡ = ---- represents a deviation from ideal solution behavior

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8. Fugacity coefficient relationships from PVTN EOS property models There are two basic approaches— one involving pressure explicit EOSs like the PR or RKS, and the other for volume explicit EOSs, such as Virial or Corresponding States formulations, including compressibility expansions in density or molar volume or similarly structured equations. Equations (9-142) and (9-143) provide convenient forms for pressure explicit EOS models for mixture and pure components, respectively, while Eqs(9-129) and (9-127) work for volume explicit EOS models. For example, for component i in a mixture:

[ ]

ln ln j

V

ii T ,V ,N i

P RTˆRT dN V∞

⎡ ⎤⎡ ⎤∂⎢ ⎥φ = − − −⎢ ⎥∂⎢ ⎥⎣ ⎦⎣ ⎦∫ V RT Z

or

0

ln P

i iRTˆRT VP

dP⎡ ⎤φ = + −⎢ ⎥⎣ ⎦∫

At this point you should know how to calculate the fugacity or fugacity coefficient for a pure component using a pressure- or volume-explicit EOS or for a component in a binary mixture using a suitable PVTN EOS that has been properly formulated with mixing rules for its constants (for example, the amix and bmix constants in the RKS EOS) in terms of pure component properties and a binary interaction parameter, eg δij. Being able to do this provides a powerful tool for calculations required later in the course, for example to estimate the vapor pressure of a pure component you would equate iφ for the liquid and vapor phases at a given temperature by estimating the P and using the EOS to calculate liquid and vapor volumes (densities) until the fugacity coefficients for each phase were equal. 9. Activity coefficients from ∆GEX property relationships The activity coefficient is defined in terms of the partial molar excess Gibbs free energy of mixing and can be directly related to the fugacity using the definition of activity:

[ ]

ln j

EXEXii

i T ,P ,N i

GRT GN

⎛ ⎞∂∆γ = ∆ = ⎜ ⎟∂⎝ ⎠

where i i i iˆ ˆa x f f x

+γ ≡ = i

To evaluate the partial derivative we need an expression for ∆GEX:

( ), , 1,...,EX IDmix iG G G f T P N i n∆ ≡ ∆ − ∆ = =⎡ ⎤⎣ ⎦

which represents the difference between the actual enthalpy of mixing and the enthalpy of mixing for an ideal solution at the same T, P and composition as the actual mixture.

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Typically, one has access to a model that gives ∆GEX = ∆GEX / N as a function of T, P, and xi, and with Eq. (9-53) one can easily obtain γi for each component, e.g.

[ ]

ln EX

EXi j

j i i T ,P ,x j ,i

GRT G xx≠

⎛ ⎞∂∆γ = ∆ − ⎜ ⎟∂⎝ ⎠

∑

In addition, the Gibbs-Duhem equation can be applied to calculate the other activity coefficient given a set of data for one component (e.g. if γsolvent is known as a function of composition then γsolute can be estimated by integrating the Gibbs-Duhem relationship). Conversely, if both activity coefficients of a binary mixture are known then the Gibbs-Duhem equation can be used to check the thermodynamic consistency of a given set of data. For example, the slope and area tests have been developed specifically for this purpose (see pages 357-358 and Fig. 9.3). A key issue here is how to deal with the standard or reference state condition (+), as that will have a direct effect on the magnitude and behavior of γi. If a symmetric reference state is used then ( i i

ˆ )f f pure i+ = and the Lewis and Randall rule is followed as xi goes to 1.0 with γi approaching 1.0. Alternatively, a unsymmetrical reference state can be used where the infinite dilution behavior as defined by Henry’s Law determines that γi

** approaches 1.0 as xi goes to 0. This is frequently called the McMillan-Mayer reference condition. Another popular alternative commonly employed for electrolytes is to use a 1 molal solution at T and P of the mixture where the mixture follows Henry’s Law in dilute solutions as xi or mi →0 (see Fig. 9.9 for example). The discussion in the text from pages 360 to 365 should be carefully reviewed to see how the activity coefficient is related to fugacity behavior for different reference state conditions. Particular attention should be paid to Figures 9.6 – 9.9.

10. Ideal reversible work associated with mixing or separating components Under isothermal conditions the reversible work is equal to the net change in Gibbs free energy associated with the process, more generally the reversible work can be related directly to the change in availability using the methods introduced in Chapter 14. Examples 9.8 and 9.9 should be reviewed to see how the concepts are employed. The approach is quite straightforward if you are dealing with a steady state process. Things are a bit more complicated if the system or process conditions are changing with time, but the general concept remains unchanged.

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':"

.

Supplementary Notes for

Chapter 14 Energy and Power Production,

Conversion, and Efficiency

1. Fundamental principles -energy conservation and the 1 st Law of thermodynamics

-entropy production and the 2nd Law of thermodynamics -reversible Carnot heat engines -maximum work I availability I exergy concepts

2. Efficiencies -mechanical device efficiency for turbines and pumps -heat exchange efficiency -Carnot efficiency ..

-cycle efficiency -fuel efficiency -utilization efficiency

" 3. Ideal cycles

-CarnQt with fixed T Hand T c -Carnot with variable T H and fixed T c -Ideal Brayton with variable T Hand T c

4. Practical power cycles -an approach to Carnotizing cycles -Rankine cycles with condensing steam or organic working fluids

-sub and supercritical operation -feed water heating -with reheat

-Brayton non-condensing gas turbine cycles -Combined gas turbine and steam Rankine cycles -Topping and bottoming and dual cycles -Otto and diesel cycles for internal combustion engines

5. Examples of power conversion using a natural gas or methane energy source -sub-critical Rankine cycle -gas turbine open Brayton cycle -combined gas turbine steam Rankine cycle -electrochemical fuel cell



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-, .."

..

For further information, refer to:

1. Milora, S.L. and Tester, J. W. Geothermal Energy as a Source of Electric Power. Cambridge, MA: MIT Press, 1976, especially chapters 3-5.

2. Balje, C.E. Turbomachines. New York: John Wiley and Sons, 1981. , 3. Balje, C.E. Journal of Engineering for Power, AS ME Transactions 84(1), 83,

January 1962.

Power Cycles

1. Rankine Cycle Limitations

-utilization vs. cycle efficiency (11u vs. 11c)

-"Carnotizing" to approach 2nd law limit of performance

t , 11 -mechanical component efficiencies 11 ~ < 1 for turbines and feed pumps (effect of moisture to decrease efficiency)

-heat transfer irreversibilities (AT> 0 in primary heat exchanger and condenser)

-materials limitations (metallurgical limit for steel in steam Rankine cycle 600°C (1100°F)

2. Improvement to Rankine Cycle (fossil or nuclear-fired)

-reheat

-supercritical vs. subcritical operation with steam

-decrease turbine exhaust pressure/condensing temperature

r!!~r;'.~":



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--~--~-~-- _f~::':'"

.

-regenerative feed water heating/interstage moisture extraction

-topping and bottoming cycles using non-aqueous fluids (topping -Hg, Cs, K; bottoming -NH3, halocarbons)

-combined cycles (gas turbine cycle linked to steam cycle)

3. Power Generation with Low Temperature Heat So~es (solar, geothennal, etc.)

-cycle configurations possible

-analysis of single and multi-single flash systems

-single (binary), sub- and supercritical cycles using non-aqueous working fluids

-derived thermodynamic property estimation from BOS, P/iquid, p:;~ and C; correlations

-effect cycle pressure on performance with an R-115 and a 150°C resource

-irreversibility analysis of performance as function of turbine inlet pressure

-11u vs. temperature for several fluids

-correlation of "degree of superheat" for optimal performance vs. C;lR

4. Thennodynamic Anaiysu of Fluid Flow in a Duct or Nozzle

-thermodynamic analysis of fluid flow in a duct or nozzle

-conversion of KE into rotating shaft work

-sonic limitations in choked flow (pressure ratio, isentropic AH)

5. Turbine, Pump and Compressor Sizing and Perfonnance

-Balje analysis of performance (11 = ffNs' Ds, Re, Mal)

-generalized approach to turbine exhaust and requirements

-sizing figure of merit

2



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\':~~;~r:;!_- ?~~:

.

Power Cycle Terminology

I ffi .net work W net 11 C = cyC e e Clency = primary heat exchanged = - QH

net work Wnet Wnet11U ==-=maximum possible work W max A B

AB = availability change = ill -T oAS = W max

To = ultimate sink temperature for heat rejection

Po = ambient pressure

For emmpie, for geothermal .systems, at steady state

I Tgf'P gf AB = AH-ToAS

To'P 0

where T gf = geothermal fluid inlet temperature P gf = geothermal fluid inlet pressure

As the cost of producing the geothermal fluid (drilling wells, etc.) increases relative to the cost of the power conversion equipment itself (heat exchangers, turbines, pumps, etc.), cycle operation at conditions approaching max 11u is favored. See Chapters 3 and 4 of Milora and Tester (1976) for further discussion.

3



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" ""c ,'::, ,~!~~~~~

~ '~'~?:

..f: ...; T E ":;...:.:::::::., HO 5Y5T M ..;;:::

BOILER

t

.I

Wp.- _.w~

CONDENSATERETURN PUMP r T'-'.~S.I 'E

C'ONCt;NSER

--

.:;i:~::.. COLD SYSTEM ':::;:~:i:~:~~: '.::;:

: ,.

Figure 1. General Rankine cycle schematic.

"

c

200

R-717(NH3)

15

-~IOO

(.)

*1...

50

00 5 10 15 20

*YCp/R=

Y-I

Figure 8. Generalized correlation for the degree of superheat above thecritical temperature for optimum utilization of geothermal fluidavailability as a function of ideal gas state reduced heat capacity.

Ws

Qh

Qc

Wp

Hot system

Cold system

Condensate return pump

General Rankine Cycle Schematic

Condenser

Boiler

ll, Mi i i ions. i i ll i

Turbine

Adapted from Tester, J. W. and Mode chael. Thermodynam cs and Its Appl catUpper Saddle R ver, NJ: Prent ce Ha PTR, 1997, p. 606, F gure 14.7.

Image by MIT OCW.



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.

..\

1000

900SATURATED STEAM, HEAT REJECTIONTEMPERATURE = 26.7oC (BOOF)

.I

BOO

700

'"Ci~

~ 600I

~~~c:0~-J 500:J...wV):J

~ SATURATED STEAM, HEAT

~ 400 REJECTION TEMPERATUf~E = ~C.90C

-(120OF)x41:~

300

SATURATED WATER, HEAT REJECTION200 TEMPERATURE = 26.7oC (60°F)

100

SATURATED WATER,HEAT REJECTIONTEMPERATURE #

0 4B.90C f1200FI

0 50 100 150 200 250

GEOTHERMAL FLUID TEMPERATURE (OCI

Figure 2. Availability or maximum useful work as a function of geothermalfluid temperature.

" ""c ,'::, ,~!~~~~~

"

c

200

R-717(NH3)

15

-~IOO

(.)

*1...

50

00 5 10 15 20

*YCp/R=

Y-I


Max

imum

use

ful w

ork

(kW

-sec

/kg)

1000

900

800

700

600

500

400

300

200

100

0 0 50 100 150 200

Geothermal fluid temperature (oc)

fluid temperature.

Saturated steam, heat rejection temperature

26.7oc (80oF) 48.9oc (120oF)

26.7oc (80oF) 48.9oc (120oF)

Availability or maximum useful work as a function of geothermal

Saturated water, heat rejection temperature

Image by MIT OCW.

250



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" ""c ,'::, ,~!~~~~~

"

c,.,

. ,

J

1 SUPERCRITICAL RANKINE CYCLE)

1..

B

l.LJa:::)

~a:l.LJ

, C-

~ NGI- 0

CO

ENTHALPY

GEOFLUIO TURBINE

B C COOLI NG

WATER OR AI R ,HEATEXCHANGER 0

CONDENSER/DESUPERHEATER

A E

WORKING ED PUMPFLUID

PRODUCTION ~WELL PUMP

REINJECTION WELL

/(

Figure 3. Supercritical Rankine cycle operating state points on a temperature-enthalpy (T-H) diagram.

200

R-717(NH3)

15

-~IOO

(.)

*1...

50

00 5 10 15 20

*YCp/R=

Y-I


Geofluid

Geofluid

Heat exchanger Condenser Desuperheater

Production well

Reinjectionwell

fluid Feed pump

pump

fluid

Cooling water or air

Cooling wateror air

Supercritical Rankine Cycle

Enthalpy

A C

D

B

E

B C

EA

Supercritical Rankine cycle operating state points on a temperature-enthalpy

D

Working

Working

Turbine

Tem

pera

ture

(T-H) diagram.

Image by MIT OCW.



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" ""c ,'::, ,~!~~~~~

c:';;':

; , .~;~7'"

.

",.

.' .'

" ..,., ..I, f .'. " 9 AT

I,," Tgf-L\T, i "

.

t :..TurbineOJ

=' .expanslon-.c...Q)

.Co ..'. '. .E ....Q) TO+6lj+h.T2 T.-

"~ .Heat rejection .6Tz ~ Qout ..

TO L-ot) ..Coolant

.

Entropy ~

Figure 4. Temperature-entropy (T -S) plot for an idealized or "Carnotized"power cycle. Note that as AT 1 and AT 2 go to zero that maximumwork output is achieved.

"

c

200

R-717(NH3)

15

-~IOO

(.)

*1...

50

00 5 10 15 20

*YCp/R=

Y-I


Tgf - T1

T1

T1

T2

T0 + T1 + T2

Coolant

Qout

Qin

Heat rejection

T0

Tgf

Entropy

Geothermal fluid cooling path

Tthat as T1 and T2 go to zero that maximum work output is achieved.

Supercritical heating path

Turbine expansion

Tem

pera

ture

emperature-entropy (T-S) plot for an idealized or "Carnotized" power cycle. Note

Image by MIT OCW.



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!~!~~~

.~ c...",.

C2FSCIMW = 154.5 g/mol

I Pc = 458 psia (32 bar) llt = 85% dry

T = 8GoG (175 goF ) IIp = 80%c .R-115 Chloropentofluoroethane

Tgt=150.CT. =26.7. C

16 @ @P=27.5 bar (398.9 psio) p. 39.26 bar (569.4 pslo)

14 F=0.87 Pr=I.24

'Iu=46.5 % '1.=56.5 %

'Icycl.=90o;. _112.'12 'ICYCII-' ,.

1

8

6I

4

u2.!!

I .20~g-16 @4J

I- P=114.4 bar (1659.2 psla)

14 P, =3.62'1.=54.6 %

12 'Icyci. =10.6 %

1

8

6

4

2

200Enthalpy (JIg)

Figure 5 Approach to thermodynamically optimized Rankine cycle for R-115with a 150°C liquid geothermal fluid source and heat rejection at26.7°C (80°F). Temperature-enthalpy (T -H) diagrams shown atdifferent reduced cycle pressures.

" ""c ,'::, ,~!~~~~~

"

c

200

R-717(NH3)

15

-~IOO

(.)

*1...

50

00 5 10 15 20

*YCp/R=

Y-I


EE

Liquid Heating

Condensation

Pumping D

C' C

B

A

Evaporation

Superh

eat

15oC

10oC

160

140

120

100

80

60

40

20

0

160

140

120

100

80

60

40

20

0 50 100 150 200

A B

C D

Tgf = 15.0o C To = 26.7o C

Pr = 0.87 u = 46.5%

t = 85% dry p = 80%

cycle = 9.0%

Condensation

Pumping D

C

B

A

Pr = 1.24 u = 56.5% cycle

Condensation

Pumping E D

C

B

A

Pr = 2.54 u = 63.2% cycle

50 100 150 200

Condensation

Pumping E D

B

A

Pr = 3.62 u = 54.6% cycle = 10.6%

o C)

Enthalpy (J/g)

C2F5Cl

PC = 458 psia (32 bar) TC = 80oC psia (175.9oF)

MW = 154.5 g/mol

oC liquid geothermal fluid source and heat rejection at 26.7oC (80o

Geothermal Fluid Cooling Path Supercritical Heating Path

Sensible Heat Rejection

ll, Mi l. i i ions i ill i

P = 27.5 bar (398.9 psia) P = 39.26 bar (569.4 psia)

= 11.2%

P = 80.1 bar (1161.8 psia)

= 11.9%

P = 114.4 bar (1659.2 psia)

Tem

pera

ture

(

R-115 Chloropentafluoroethane

Approach to thermodynamically optimized Rankine cycle for R-115 with a 150F). Temperature-enthalpy (T-H)

diagrams shown at different reduced cycle pressures.

Turbine Expansion

Adapted from Tester, J. W. and Mode chae Thermodynam cs and Its Appl cat . Upper Saddle R ver, NJ: Prent ce Ha PTR, 1997, p. 49, F g. 14.10.

Image by MIT OCW. www.bsscommunitycollege.in www.bssnewgeneration.in www.bsslifeskillscollege.in


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" ""c ,'::, ,~!~~~~~

"

c

200

R-717(NH3)

15

-~IOO

(.)

*1...

50

00 5 10 15 20

*YCp/R=

Y-I


12

10

9

8

7

6

5

4

3

2

1

0 0

2

4

6

8

10

12

14

16

18

20

65

60

55

50

45

40

1.0 1.5 2.0 2.5 3.0 3.5 4.0 5.0

u , %

of

B

I, S

peci

fic ir

reve

rsib

ility

(J/g

)

I T

Pr

Component irreversibility and u as a function of reduced cycle pressure

oC geothermal fluid temperature To = 26.7oC condensing temperature

ll, Mi l. i i ionsRi i ll i

Utilization factor

I Heat rejection I Residual fluid

I Pump

I Heat exchanger

11

0.8 0.9

, Tot

al ir

reve

rsib

ility

(J/g

)

, reduced cycle pressure

for R-115 for the condition shown in the figure above.

R - 115 chloropentafluoroethane T = 150

Adapted from Tester, J. W. and Mode chae Thermodynam cs and Its Appl cat . Upper Saddle ver, NJ: Prent ce Ha PTR, 1997, p. 613, F g. 14.11.

I Total

I Turbine

Image by MIT OCW.



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" ""c ,'::, ,~!~~~~~

"--~~:fi

.

6

6

-~!...5

~~>-u5Z~ .R- 717 (NH3)~ ..R -115lJ..lJ.. 4 .R -32w 0 R- 22Z0 6. RC -318~ 4 x R -114

~ .R -600 a (ISOBUTANE)~§ 3 OPJIMUM THERMODYNAMIC PERFORMANCE

T = 16.7°C T = 267°C0 cond.

3 TI, = 85% TIp = 80%

~Tplnch = 10°C

150 200 250 300

GEOTHERMAL FLUID TEMPERATURE (OC)

Figure 7. Utilization efficiency 11u as a function of geothermal fluid (initialheat source) temperature for optimum thermodynamic operatingconditions.

c

200

R-717(NH3)

15

-~IOO

(.)

*1...

50

00 5 10 15 20

*YCp/R=

Y-I


Util

izat

ion

effic

ienc

y u

(%)

65

60

55

50

45

40

35

30

To = 16.7oC t =

Tcond = 85 % p = 80 %

Optimum Thermodynamic Performance 26.7oC

Tpinch = 10oC

100 150 200 250 Geothermal fluid temperature (oC)

Utilization efficiency u as a function of geothermal fluid (initial heat source) temperature for optimum thermodynamic operating conditions.

R - 717 (NH3) R - 115 R - 32 R - 22

RC - 318 R - 114 R - 600a (Isobutane)

Image by MIT OCW.

300



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" ""c ,'::, ,~!~~~~~

"

c

200

R-717(NH3)

15

-~IOO

(.)

*1...

50

00 5 10 15 20

*YCp/R=

Y-I


200

150

100

50

0 0 5 10 15 20

T* -

T c ( o C

)

=

- 1

R - 717 (NH3)

R - 32

R - 600a

R - 22

R - 114

R - 115

RC - 318

Generalized correlation for the degree of superheat above the critical temperature for optimum utilization of geothermal fluid availability as a function of ideal gas state reduced heat capacity.

C* /R 790

p

C* P /R

Image by MIT OCW.



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Supplemental Notes for Chapter 17 General Treatment of Phase and Chemical Equations

1. Phase rule constrained parameter variability - F = n + 2 - π -r

- invariant (F = 0) - monovariant (F = 1) - divariant (F = 2)

- other constraints (stoichiometric ratio, “indifferent states”) 2. Matrix/determinant formalism applied to generalized Gibbs-Duhem and reaction equilibrium

expressions

- phase and chemical equilibrium criteria combined - Gibbs-Duhem combined with reaction equilibrium criteria to produce a generalized

expression for π phases, r reactions and n components

( ) ( ) ( ) 1 1

∑ ∑π

= =⎥⎦

⎤⎢⎣

⎡µ−+−

s

n

ii

si

ss dNdPVdTS and ( )∑ ∑= =

µνn

i

r

ki

ki d

1 1

( )

( )

( )

( )

[ ]

( )

( )

( )

( )

[ ]

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

0

0

0

0

0

1

1

1111

1

1111

1

1

1

1

=

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−π

n

j

rn

rj

r

nj

πn

πj

π

nj

r

π

rdµ

dµ

dµ

ννν

ννν

NNN

NNN

dP V

V

dT S

S

or in shorthand vector notation

( ) ( ) ( ) 0=µ−+− js

jss dNdPVdTS




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3. Invariant systems (F = 0)

- dµi = 0 - mass and mole constraints - | N (s) | ≠ 0 required or the rank of the N (s) matrix must be reduced to avoid an over-

constrained system - indifferent states result if | N (s) | = 0

4. Monovariant systems (F = 1) pressure-temperature variations - generalized Clapeyron equation - use Gibbs-Duhem with chemical reactions included - use a chemical potential or fugacity approach to get

1

∆HdPdT T ∆V=

=F

or equivalently V∆S∆

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

( )

( )

( )

( )

11 1 11 2

1 2

1 1 11 2

1 2

0

0

n

ππ π πn

n

r r rn

x H x x

x H x x∆H

ν ν ν

ν ν ν

≡

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

( )

( )

( )

( )

11 1 11 2

1 2

1 1 11 2

1 2

0

0

n

ππ π πn

n

r r rn

x V x x

x V x x∆V

ν ν ν

ν ν ν

≡

5. Isobaric monovariant equilibria, temperature-composition variations

-apply dP = 0 constraint to the set of generalized Gibbs-Duhem equations and

use the Gibbs-Helmholtz relationship to simplify

[ ]

( ) ( )( )

⎥⎦⎤

⎢⎣⎡ −

∂∂−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

∂

∂s

iβ

T,Pββs

i

,πα,P,β

xH∆H

x/µxT

xT

1

11

1 …




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or with the fugacity introduced

[ ]

( )

( )⎥⎦⎤

⎢⎣⎡ −∆

⎥⎥⎦

⎤

⎢⎢⎣

⎡

∂

∂−

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

∂

∂β

β

β

παβ s

i

PT

s

P xHH

xf

xRT

xT

1

,1

11

2

,,,1

ˆln

…

6. Indifferent states and azeotropic behavior (F ≥ 2) ( )s

jN or ( ) 0sjx =

- rank of the mole or composition matrix must be reduced which is equivalent to reducing the number of components

- Gibbs-Konovalows’ first and second theorems ( ) ( ) ( ) 0=µ−+− j

sj

ss dNdPVdTS

if ( )sjN = 0 and if dP = 0 then dT = 0 (extremum in T)

or if ( )s

jN = 0 and if dT = 0 then dP = 0 (extremum in P)

( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

ννν

ννν=

πππ

rn

r

nj

nj

nj

sj

NNN

NNN

N

rj1

1111

1

1111

and

( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

ννν

ννν=

π

πππ

nrj

r

nj

nj

nj

sj

xxx

xxx

x

1

1111

1

1111




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10.40 AppendixConnection to Thermodynamics and Derivation of

Boltzmann Distribution

Bernhardt L. Trout

Outline

• Cannonical ensemble• Maximum term method

• Most probable distribution• Ensembles continued: Canonical, Microcanonical, Grand Canonical, etc.• Connection to thermodynamics• Relation of thermodynamic quantities to Q

0.1 Canonical ensemble

In the Canonical ensemble, each system has constant N,V ,and T.After equilibration, remove all of the systems from the bath, and put them all

together:Apply postulate 2 to the ensemble of systems, also called a supersystem.Let nj = number of systems with energy Ej . Also, N =

Pj nj and Etot =P

j njEj .If we know all Ej ’s, then the state of the entire ensemble would be well-defined.

For example, let’s analyze an ensemble with 4 systems, labeled A, B, C, and D,where

A B C DE2 E3 E2 E1

Then, Etot =E1 + 2E2+E3Also, the distribution of the systems, −→n = (n1, n2,n3, ...) = (1, 2, 1).But there are many different supersystems consistent with this distribution. In

fact, the number of supersystems consistent with this distribution is

Ωtot(−→n ) = N !Q

j nj !=

4!

1!2!1!= 12

What is the probability of observing a given quantum state, e.g. Ej? In otherwords, what is the fraction of systems in the ensemble in the state Ej?

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10.40: Fall 2003 Appendix 2

N,V,EA

N,V,Ej

N,V,EB N,V,EC

. . . . . .

(Bath at temperature T)

(Ensemble of systems)

Figure 1: Canonical ensemble

The answer is njN .

However, it may be the case that many distributions fulfill the conditions of theensemble, (N,V ,Etot).

For example, assume that there are two:

n1 = 1, n2 = 2, n3 = 1;Ωtot = 12

and

n1 = 2, n2 = 0, n3 = 2;Ωtot = 6

Then the probability of observing, for example, E3 is 14 in the first distributionand 1

2 in the second.The probability in the case where both distributions make up the ensemble is:

p3 =

µ1

4

¶1× 12 + 2× 6

12 + 6=1

3

In general,

pj =

µ1

N¶ P

−→n Ωtot(−→n )nj(−→n )P

−→n Ωtot(−→n )

where the sum is over all distributions satisfying the conditions of (N,V ,Etot).Then, for example, we could compute ensemble averages of mechanical quanti-

ties:E = hEi =

Xj

PjEj

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N,V,EAN,V,Ej

N,V,EB N,V,EC. . . . . . Figure 2: Canonical ensemble forming its own bath

andP = hP i =

Xj

pjPj ,

where p is the pressure.

0.2 Maximum term method

Recall:

pj =

µ1

N¶ P

−→n Ωtot(−→n )nj(−→n )P

−→n Ωtot(−→n )

where

Ωtot(−→n ) = N !Q

j nj !.

As N →∞, nj →∞, for each j.Thus, the most probable distribution becomes dominant. We can call this dis-

tribution, −→n ∗.Let n∗j = nj in the

−→n ∗ distribution. Then

pj =1

NΩtot(

−→n ∗)n∗jΩtot(

−→n ∗) =n∗jN

0.3 Most probable distribution

Which distribution gives the largest Ωtot?Solve via method of undetermined multipliers:Take natural log of Ωtot.

ln (Ωtot(−→n )) = ln

µ N !Qi ni!

¶=

ÃXi

ni

!ln

ÃXi

ni

!−Xi

ni lnni,

where we have switched the index from j to i and used Stirling’s approximation,which becomes exact as ni →∞ :

ln y! ≈ y ln y − y.

We wish to find the set of nj ’s, which maximize Ωtot(−→n ) and hence ln(Ωtot(−→n )) :

∂

∂nj

"ln (Ωtot(

−→n ))− αXi

ni − βXi

niEi = 0

#, j = 1, 2, 3, ...

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where α and β are the undetermined multipliers. Carrying out the differentiationyields

ln

ÃXi

ni

!− lnn∗j − α− βEj = 0, j = 1, 2, 3, ...

orn∗j = N e−αe−βEj , j = 1, 2, 3, ...

Recalling thatN =

Xj

n∗j

yields Xj

e−αe−βEj = 1

oreα =

Xj

e−βEj .

Also,

hEi =P

j n∗jEj

N =

Pj N e−αe−βEjEj

N =

Pj e−βEjEjPj e−βEj

and

pj =n∗jN = e−αe−βEj =

e−βEjPj e−βEj ,

whereQ =

Xj

e−βEj

and, as we discussed in the last lecture, is the partition function, the normalizationfactor.

0.4 Canonical ensemble continued and connection to ther-modynamics

Recall from last time, via the maximum-term method in the canonical ensemble:

hEi =P

j n∗jEj

N =

Pj N e−αe−βEjEj

N =

Pj e−βEjEjPj e−βEj

and

pj =n∗jN = e−αe−βEj =

e−βEjPj e−βEj

,

where,Q =

Xj

e−βEj ,

as we discussed in the last lecture, is the partition function, the normalization factor.

In addition, as we have shown:

E = hEi =Xj

pjEj

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andP = hP i =

Xj

pjPj ,

where P is the pressure.

If we differentiate the equation for hEi,

dhEi =Xj

Ejdpj +Xj

pjdEj

= − 1β

Xj

(ln pj + lnQ) dpj +Xj

pj

µ∂Ej

∂V

¶N

dV . (1)

Recall that the pressure,

P =

µ∂E

∂V

¶N

or

Pj =

µ∂Ej

∂V

¶N

.

This yields for equation 1

dhEi = − 1β

Xj

ln pjdpj − 1β

Xj

lnQdpj +Xj

pjPjdV .

[Note that

d

Xj

pj ln pj

=Xj

ln pjdpj +Xj

pjd (ln pj)

=Xj

ln pjdpj +Xj

pjdpjpj

. (2)

Since, Xj

pj = 1,

Xj

dpj = 0.]

Thus, the right term in equation 2 is equal to 0. This yields

dhEi = − 1βd

Xj

pj ln pj

− hP idV .Recalling from the combined first and second laws (in intensive form, noting thatsince N is a constant, intensive and extensive forms are equivalent):

dU = TdS − PdV

Since U ↔ hEi and p↔ hpi,

TdS ↔ − 1βd

Xj

pj ln pj

.

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LetX = −

Xj

pj ln pj .

Then,

dS =1

βTdX. (3)

We know that the left side of the equation is an exact differential, so the right sidemust be too, and thus, 1

βT must be a function of X. This means that

dS = φ(X)dX = df(X).

Integrating,S = f(X) + const, (4)

where we can set the arbitrary constant, const, equal to 0 for convenience.Now we can make use of the additive property of S, and we can divide a system

into two parts, A and B. This yields:

S = SA + SB = f(XA) + f(XB). (5)

Note thatXA+B = −

Xi,j

pi,j ln pi,j ,

where i is the index for the possible states of A and j is the index for the possiblestates of B. Then

XA+B = −Xi,j

pAi pBj (ln p

Ai + ln p

Bj )

= −Xi

pAi ln pAi −

Xj

pBj ln pBj

= XA +XB.

Thus, from equation 5,

S = f(XA) + f(XB) = f(XA +XB).

For this to be so,f(X) = kX,

where k is a constant. Thus,

S = −kXj

pj ln pj . (6)

From equations 3 and 4,1

βT= k,

and thus,

β =1

kT.

We designate k as Boltzmann’s constant, a universal constant.

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0.5 Microcanonical, Grand Canonical, and other ensembles

Recalling the formulation for S from equation 6, and noting that in the microcanon-ical ensemble,

pj =1

Ω,

where we recall that Ω is the total number of states with the same energy, then

S = −kXj

pj ln pj = −kXj

1

Ωln1

Ω

= k lnΩ(N,V ,E).

This is Boltzmann’s famous formula for the entropy.In the Grand Canonical ensemble, the number of particles in each system is

allowed to fluctuate, but µ is kept constant. This is called the (V , T, µ) ensemble.Also, there are other ensembles, such as (N,P, T ), etc. Note that from an analysisof fluctuations (Lecture 27), we shall see that in the macroscopic limit of a largenumber of systems, all of these ensembles are equivalent.

0.6 Relation of thermodynamic quantities to Q

Recall that

S = −kXj

pj ln pj

pj =e−βEj

Q

Q =Xj

e−βEj

Plugging in the formula for pj into that for S yields

S = −kXj

e−βEj

Qln

e−βEj

Q

= −kXj

e−βEj

Q

µ−Ej

kT− lnQ

¶=

hEiT+ k lnQ

Recalling our definitions from macroscopic thermodynamics and the fact that U ↔hEi yields

A= −kT lnQSimilarly,

S = −µ∂A

∂T

¶V ,Ni

= kT

µ∂ lnQ

∂T

¶V ,Ni

+ k lnQ

P = −µ∂A

∂V

¶T,Ni

= kT

µ∂ lnQ

∂V

¶T,Ni

U = A+ TS = kT 2µ∂ lnQ

∂T

¶V ,Ni

.

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Thus, all thermodynamic properties can be written in terms of the partition func-tion, Q(N,V , T )!In order to compute Q, all we need are the possible energy levels of the system.

We can obtain these from solving the equations of quantum mechanics.

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First Postulate:For closed simple systems with given internal restraints, there

exist stable equilibrium states that can be characterized completely by two independently variable properties in addition to the masses of the particular chemical species initially charged.

• "closed, simple systems"• "independently variable properties"• masses of each component known• consistent with the Gibbs phase rule• restatement of Duhem's theorem

L - Vcoexist


P

T

L

S

Crit. Pt.

L

V

Crit. Pt.

P

V

S

S+V

L

L+VS+LS

S+V

L

L+VS+L



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10.40 Lecture 21Fundamental Princicples of Quantum and

Classical Statistical Mechanics

Bernhardt L. Trout

October 14, 2003

(In preparation for Lecture 21, also read T&M, beginning of Ch. 10 through10.1.3).

Now, we turn to the microscopic: atoms and molecules. Why do we need touse the theory of atoms and molecules and how do we know that they really exist?

(Think about these issues for class discussion.)

Concurrent with the development of the atomic theory is the concept of statisti-cal averaging. Thermodynamic properties, here known as macroscopic properties,such as T,P,U, etc. are the result of averaging the properties of many differentatoms and molecules. This is useful, because the important property, the micro-scopic property, of each atom or molecule is its energy. Averaging the energies ofa large number of atoms and/or molecules in the proper way will yield the macro-scopic properties of the system. Thus, the need to average introduces the need forstatistics and for probability.

21.1 Objectives of 10.40 Part II

The objectives of this part of the course is to give you an appreciation of the toolsof statistical thermodynamics and statistical mechanics and how they can be used(1) to compute thermodynamic quantitaties for specific systems and (2) tobetter understand thermodynamic quantitites in general. It is not expectedthat you will have a full understanding of statistical thermodynamics and statisticalmechanics. You should, however, be able to solve simple problems and understandenough to know how to increase your understanding, if so needed for your research.

21.2 Quantummechanics yields the energy levels of a system

The main equation of quantum mechanics, the Schrödinger equation, gives thepossible energy levels of a system:

H|ψνi = Eν |ψνi,where ν = 0, 1, ....ψν is the state of the system and Eν is the energy of state ν. ψ0is the ground state of the system, ψ1 is the first excited state, etc. An example canbe seen in Figure 1.

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10.40: Fall 2003 Lecture 21 2

.

.

.

E0E1

.

.

.

Figure 1: Example of different energy levels coming from the solution toSchrödinger’s equation. Note that it does not matter if we write E in intenstive orextensive form. Only the units will change, e.g. kJ versus kJ/mol.

The thermal energy is in units of kT (k is Boltzmann’s constant = 1.381021×10−23 J K−1; R = kNA), where 1 kcal mol−1= 505 K = 351 cm−1.If one were to solve Schrödinger’s equation, for example, for you in 66-110, and

treating you as a particle in a box, if you weigh 60 kg, then for you, E1 − E0 ≈1× 10−50 kJ mol−1.Note that quantum mechanics does not in itself tell which energy levels are

occupied and the degree of occupancy. Those are in the realm of statistical me-chanics. In this sense, the solutions to the Schrödinger equation are only thepossible energy levels.

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10.40: Fall 2003 Lecture 21 3

21.3 Probabilities and averaging

An observable property, Bobs, is some mechanical, measured property, that is anaverage over many measurements of short time:

Bobs =1

Nm

NmXj=1

Bj ,

where Nm is the number of measurements.

Examples of mechanical properties: V, P,N,E,M (M is the magetization.)Examples of non-mechanical properties: T, S,A, µ

If pj is the probability that the system is in state j, then

Bobs =Xj

pjBj = hBi,

where < B > is called the ensemble average, and an ensemble is a collection ofmicrostates (see below). Also note, thatX

j

pj = 1.

In systems at equilibrium, it turns out, as we shall show later,

pj =e−βEj

Q,

where Ej is the energy of state j, β = 1/kT, and

Q =Xj

e−βEj ,

where Q is the partition function. It is merely the normalization factor, so thatPj pj = 1.

e−βEj is called the Boltzmann factor.

21.4 Example of application of the Boltzmann factor

Let us analyze a two-state system, where E1 −E0 = 1.0 kcal mol−1. At any giventime, what is the probability to find the system in the ground state at 300 K andat 1000 K.At 300 K, kT = 0.6 kcal mol−1, and at 1000 K, kT = 2.0 kcal mol−1.At 300 K,

p0 =e−E0/0.6

e−E1/0.6 + e−E0/0.6=

e−E0/0.6

e−E0/0.6e−(E1−E0)/0.6 + e−E0/0.6=

1

e−(E1−E0)/0.6 + 1= 0.84.

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10.40: Fall 2003 Lecture 21 4

Similarly, at 1000 K,

p0 =1

e−(E1−E0)/2.0 + 1= 0.62.

limT→∞

p0 =?

limT→0

p0 =?

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10.40 Lecture 22Postulates of statistical mechanics, Gibbs

ensembles

Bernhardt L. Trout

October 14, 2003

(In preparation for Lecture 22, also read T&M, 10.1.4).

Outline

• Ensembles• Postulates of statistical mechanics• Gibbs Ensembles: Microcanonical ensemble• Gibbs Ensembles: Canonical ensemble• Gibbs Ensembles: Grand Canonical and othersNote that the derivations of the relationships between the thermodynamicquantities and the statistical quantities are presented in the Appendix.

22.1 Ensembles and ensemble averages

An ensemble is an assembly of microstates or systems. Imagine a very large collec-tion of systems evolving in time. A snapshot of the state of each of these systemsat some instant in time forms the ensemble.

22.2 Two postulates of statistical mechanics:

1. time averaging is equivalent to ensemble averaging

2. for possible states with the same N,V , and E, all states are equally likely.Atkins: ”principle of equal a priori probabilities”

22.3 Gibbs Ensembles: Microcanonical ensemble

In the microcanonical ensemble, each system has constant N,V , and E.

By Postulate 2, the probability that any give system is in a particular state, j, is1

Ω(N,V ,E) , where Ω(N,V ,E) is the total number of possible states. It turns out (seethe Appendix for derivations) that the connection to macroscopic thermodynamicsis through the entropy:

S = k lnΩ(N,V ,E).

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10.40: Fall 2003 Lecture 22 2

N,V,E

N,V,E

N,V,E N,V,E. . . . . .


Figure 1: Microcanonical ensemble

22.4 Gibbs ensembles: Canonical ensemble

In the Canonical ensemble, each system has constant N,V ,and T.After equilibration, remove all of the systems from the bath, and put them all

together:

Here, the connection is through the Helmholtz Free Energy:

A= −kT lnQ(N,V , T )

Similarly,

S = −µ∂A

∂T

¶V ,N

= kT

µ∂ lnQ

∂T

¶V ,N

+ k lnQ

P = −µ∂A

∂V

¶T,N

= kT

µ∂ lnQ

∂V

¶T,N

U = A+ TS = kT 2µ∂ lnQ

∂T

¶V ,N

.

Thus, all thermodynamic properties can be written in terms of the partition func-tion, Q(N,V , T )! In order to compute Q, all we need are the possible energylevels of the system. We can obtain these from solving the equations of quantummechanics.

22.5 Gibbs Ensembles: Grand Canonical and others

In the Grand Canonical ensemble, the number of particles in each system is allowedto fluctuate, but µ is kept constant. This is called the (V , T, µ) ensemble. Also,there are other ensembles, such as (N,P, T ), etc. Note that from an analysis offluctuations (Lecture 27), we shall see that in the macroscopic limit of a largenumber of systems, all of these ensembles are equivalent. The choice of which oneto use is made for convenience.

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10.40: Fall 2003 Lecture 22 3

N,V,EA

N,V,Ej

N,V,EB N,V,EC

. . . . . .

(Bath at temperature T)


Figure 2: Canonical ensemble

N,V,EAN,V,Ej

N,V,EB N,V,EC. . . . . . Figure 3: Canonical ensemble forming its own bath

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10.40 Lectures 23 and 24Computation of the properties of ideal gases

Bernhardt L. Trout

October 16, 2003

(In preparation for Lectures 23 and 24, also read T&M, 10.1.5-10.1.7).

Outline

• Degrees of freedom• Computation of the thermodynamic properties of a monatomic ideal gas• Computation of the thermodynamic properties of polyatomic ideal gases, in-cluding diatomic ideal gases

• Summary of thermodynamic functions

23.1 Degrees of freedom

We have discussed translational, electronic, and nuclear degrees of freedom. Di-atomic and polyatomic molecules have two additional internal degrees of freedom,vibrational and rotational degrees of freedom.Excluding electronic and nuclear degrees of freedom, atoms and molecules have

a total of 3Natoms degrees of freedom, where Natoms is the number of atoms in amolecule (or atom). Linear molecules have 2 rotational degrees of freedom andnon-linear molecules have 3 rotational degrees of freedom. The rest of the degreesof freedom, 3Natoms-5 for linear molecules and 3Natoms-6 for non-linear moleculesare considered to be vibrational degrees of freedom. From now on, we approximatethese vibrational degrees of freedom as normal modes, i.e. modes of simple harmonicoscillators. We will also ignore nuclear degrees of freedom from now on.

23.2 Computation of the thermodynamic properties of a monatomicideal gas

We have N independent and indistinguishable particles. Let’s label these particles,a, b, .... Particle a can take on energy levels εaj , where as before, j is just an indexreferening each of the possible states. Particle b can take on energy levels εbj , etc.Designate the partition function of each particle with the symbol q. Thus,

qa =Xj

e−βεaj ,

qb =Xj

e−βεbj ,

etc.

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10.40: Fall 2003 Lectures 23 and 24 2

Then, the energy for the entire system, will be

Ei,j,... = εai + εbj + .... (1)

Thus, Q should be of the form Xi,j,...

e−βEi,j,... = qN ,

where q is the partition function for an individual particle.However, since each particle is indistinguishable, the ε’s in equation 1 can be

permuted in N ! ways. Thus, in order to avoid overweighting each quantum state,we much divide qN by 1

N ! . This leads to

Q = 1N!q

N .

Now, we need to compute q for a single atom. We know that from quantummechanics, the energy levels of a particle due to their translational degrees of freedomare

εj = εlxlylz =h2(l2x + l2y + l2z)

8mV 2/3,

where h is Planck’s constant, equal to 6.626×10−34 J s; lx, ly, lz = 1, 2, 3, ...; m is themass of the particle; and V is the volume occupied by the particle. The numberslx, ly, lz are the quantum numbers designating the translational quantum level. Let

R2 = l2x + l2y + l2z =8mV 2/3ε

h2 . Then, the number of states with energy < ε is

Φ(ε) =πR3

6=

π

6

µ8mε

h2

¶3/2V .

This result can be visualized by thinking of the quantum energy levels as points ona 3D Cartisian grid in an octet of a sphere.Let us remind ourselves of the difference between the designation of states and

energy levels, as shown in Figure 1. Thus,

q =X

states j

e−βεj =X

levels i

ωie−βεi ,

where ωi is the degeneracy of level i. (The degeneracy is the number of states withenergy εi.)Write

q =∞Xlx

∞Xly

∞Xlz

e−βεlxlylz .

For states with energies that are very close together, we can replaceP’s with

R’s.

How close do they need to be? How close are they? Thus,

q =

Z ∞0

ω(ε)e−βεdε,

where the number of states between ε and ε+ dε are

ω(ε)dε =dΦ

dεdε

=π

4

µ8m

h2

¶3/2V ε1/2dε.

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states energy levelsj=1

j=8j=7

j=6j=5j=4j=3

j=2

.

.

.

.

.

.

i=1

i=5

i=4

i=3i=2

Figure 1: Difference between states and energy levels.

Plugging this into the integral yields

q =π

4

µ8m

h2

¶3/2V

Z ∞0

ε1/2e−βεdε

=

µ2πmkT

h2

¶3/2V .

Thus,

q = VΛ3 ,

where Λ ≡³

h2

2πmkT

´1/2and is called the thermal deBroglie wavelength. It gives

the characteristic wavelength of a gas. Note that this q will be designated as qtbelow for "translational" (see below).From all of this, we can determine the partition function of the system,

Q =1

N !qN =

1

N !

µV

Λ3

¶N.

Now, recalling the formulas for the thermodynamic quantities in terms of Q yields

A = −kT lnQ= −kT (−N lnN +N +N ln q)

= −NkT ln

"µ2πmkT

h2

¶3/2V

Ne

#

or in intensive form,

A = −kT ln"µ2πmkT

h2

¶3/2V e

#,

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and

P = kT

µ∂ lnQ

∂V

¶T,N

= NkT

µ∂ ln q

∂V

¶T,N

=NkT

V,

and

U = kT 2µ∂ lnQ

∂T

¶V ,N

= NkT 2d lnT 3/2

dT

=3

2NkT,

and,

Cv =

µ∂U

∂T

¶V ,N

=3

2Nk.

Recalling an expression for S,

S =U −A

T

= Nk ln

"µ2πmkT

h2

¶3/2V

Ne5/2

#.

Also, H= U + PV = U +NkT and G= A+ PV = A+NkT.Next, we may need to consider internal degrees of freedom of the atom, such as

electronic and nuclear degrees of freedom. Thus, we can write the partition functionas

Q =1

N !(qtqeqn)

N ,

where qt is the translational partition function, qe is the electronic partition func-tion, and qn is the nuclear partition function. For most systems, the first excitedelectronic state and nuclear state are at ≈ 20 kcalmol−1 and 20,000 kcalmol−1

respectively. Thus, they do not play a significant role at temperatures of interest.Note that there are important exceptions to this rule of thumb, such as with thealkai metal atoms and halogens, but we will not concern ourselves with these in thiscourse.We often need to take into account degenerate electronic states of atoms that

are a result of their spins. For example, H has a net spin of 12 , meaning that it hastwo spin degrees of freedom at the ground state energy level, ωe = 2 and qe = 2.Thus an additional factor of Nk lnωe must be added to the formula for the entropygiven above, and an additional factor of NkT lnωe must be added to the formulafor the Helmholtz free energy given above.

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23.3 Computation of the thermodynamic properties of poly-

atomic ideal gases, including diatomic ideal gases

Recall thatQ =

1

N !qN .

First, we assume that all of the degrees of freedom are separable. Thus,

q(V , T ) = qt(V , T )qr(T )qv(T )qe(T ).

where qt is the translational partition function, qr is the rotational partition func-tion, qv is the vibrational partition function, and qe is the electronic partition func-tion. We discussed qt and qe last time. We note here that if we choose the electronicenergy of separated atoms as the electronic reference state, then we can define De

as the dissociation energy, the energy needed to atomize the molecule.Now, we need to compute qr and qv.

23.3.1 Vibrational degrees of freedom

From quantum mechanics, the energy levels of a harmonic oscillator are:

εn = (n+1

2)hν, (2)

n = 0, 1, 2, ...,

where h is Planck’s constant and ν is the frequency. The vibrational partitionfunction is then

qv =e−Θv/2T

1− e−Θv/T,

where Θv = hν/k. (You can verify this yourself.)Then,

Av = −NkT ln qv

= NkT

·Θv2T

+ ln³1− e−Θv/T

´¸,

Uv = NkT 2µ∂ ln qv∂T

¶= Nk

µΘv2+

ΘveΘv/T − 1

¶,

CV v =

µ∂U

∂T

¶V,N

= Nk

µΘvT

¶2eΘv/T¡

eΘv/T − 1¢2 ,and,

Sv =Uv −Av

T

= Nk/T

µΘv

eΘv/T − 1¶−Nk ln

³1− e−Θv/T

´.

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We also emphasize that the terms Θv2 in the expressions for Uv and Av above are aresult of the fact that the ground state of the harmonic oscillator has a finite energylevel as seen in equation 2. The term 1

2hν =¡kΘv2

¢is called the zero point energy

of the vibration.We note that for a diatomic molecule, for example, the measured bond dissoci-

ation energy, D0, is not the energy of atomization described at the beginning of thelecture. This is because the zero point energy adds a destabilizing contribution.Thus,

D0 = De − 12hν.

This can be easily generalized for polyatomic molecules.

23.4 Rotational degrees of freedom

From quantum mechanics, the energy levels of a rigid, linear rotator are

εj =j(j + 1)h2

8π2I,

j = 0, 1, 2, ...,

where I is the moment of inertia. Note that for a diatomic molecule consisting ofatoms 1 and 2, I = µd2, where µ is the reduced mass

³µ = m1m2

m1+m2

´and d is the

bond length. This can be generalized for polyatomic molecules. (See books onclassical mechanics.)For diatomic and non-linear polyatomic molecules, the rotational partition func-

tion is then

qr =∞Xj=0

ωje−βεj

=∞Xj=0

(2j + 1)e−j(j+1)Θr/T ,

where

Θr =h2

8π2Ik

At high T ,

qr →Z ∞o

(2j + 1)e−j(j+1)Θr/Tdj

=T

Θr=8π2IkT

h2.

Note that in order to eliminate double counting, symmetry must be taken intoaccount. The symmetry number is symbolized as σ, and for a diatomic molecule,σ = 1 if the molecule is unsymmetrical and σ = 2 if the molecule is symmetrical.A non-linear polyatomic molecule will have 3 rotational degrees of freedom, and

obviously the 3 different moments of inertia will almost always be different. Forthese species,

qr =π1/2

σ

µ8π2IAkT

h2

¶1/2µ8π2IBkT

h2

¶1/2µ8π2ICkT

h2

¶1/2,

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where σ is again the symmetry factor, which can take on many values, up to 12, andIA, IB, IC are the three principle moments of intertia. (We ignore the derivationhere, because it is somewhat complicated. It can be found in a book on classicalmechanics.)This equation can be written more compactly as

qr =π1/2

σ

µT 3

ΘAΘBΘC

¶1/2.

From this equation, we can derive the thermodynamic functions:

Ar = −NkT ln

"π1/2

σ

µT 3

ΘAΘBΘC

¶1/2#,

Er =3

2NkT,

CV r =3

2Nk,

Sr = Nk ln

"π1/2

σ

µT 3e3

ΘAΘBΘC

¶1/2#.

23.5 Summary of thermodynamic functions

Below is a summary of the thermodynamic functions, excluding nuclear and excitedelectonic degrees of freedom.

23.5.1 Monatomic ideal gas

A = −NkT ln

"µ2πmkT

h2

¶3/2V

Ne

#;

U =3

2NkT ;

CV =3

2Nk;

S = Nk ln

"µ2πmkT

h2

¶3/2V

Ne5/2

#.

Also, H = U + PV = U +NkT and G= A+ PV = A+NkT.

23.5.2 Diatomic and linear polyatomic ideal gas

− A

kT= ln

"µ2πmkT

h2

¶3/2V

Ne

#+ln

·8π2IkT

σh2

¸−3Na t om s−5X

i=1

·Θi,v2T

+ ln³1− e−Θi,v/T

´¸+De

kT+lnωe;

U

kT=3

2+2

2+

3Na t om s−5Xi=1

·Θi,v2T

+Θi,v/T

eΘi,v/T − 1¸− De

kT;

CV

k=3

2+2

2+

3Na t om s−5Xi=1

"µΘi,vT

¶2eΘi,v/T¡

eΘi,v/T − 1¢2#;

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S

k= ln

"µ2πmkT

h2

¶3/2V

Ne5/2

#+ln

·8π2IkTe

σh2

¸+

3Na t om s−5Xi=1

·Θi,v/T

eΘi,v/T − 1 − ln³1− e−Θi,v/T

´¸+lnωe.

Note that m is the mass of the molecule. Also, H = U + PV = U + kT andG = A+ PV = A+NkT.

23.5.3 Non-linear polyatomic ideal gas

− A

kT= ln

"µ2πmkT

h2

¶3/2V

Ne

#+ln

"π1/2

σ

µT 3

ΘAΘBΘC

¶1/2#−3Na t om s−6X

i=1

·Θi,v2T

+ ln³1− e−Θi,v/T

´¸+De

kT+lnωe;

U

kT=3

2+3

2+

3Na t om s−6Xi=1

·Θi,v2T

+Θi,v/T

eΘi,v/T − 1¸− De

kT;

CV

k=3

2+3

2+

3Na t om s−6Xi=1

"µΘi,vT

¶2eΘi,v/T¡

eΘi,v/T − 1¢2#;

S

k= ln

"µ2πmkT

h2

¶3/2V

Ne5/2

#+ln

"π1/2e3/2

σ

µT 3

ΘAΘBΘC

¶1/2#+

3Na t om s−6Xi=1

·Θi,v/T

eΘi,v/T − 1 − ln³1− e−Θi,v/T

´¸+lnωe.

Note that m is the mass of the molecule. Also, H = U + PV = U + kT andG = A+ PV = A+NkT.

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10.40 Thermodynamics Fall 2003 Problem Set 1

1. Problem 3.9

2. Problem 3.15 Be sure to state and justify all assumptions made

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10.40 Thermodynamics Fall 2003 Problem Set 2 1. Problem 4.3 2. You are asked to prove that ∆S for an irreversible, adiabatic process is always > 0.

Consider the following closed system cyclic process for an ideal gas.

I. An irreversible, adiabatic expansion from point 1at TA , PA to point 2 at TB, PB where TA > TB and PA > PB .

II. A reversible, isothermal compression from point 2 to point 3 at TB III. A reversible, adiabatic compression from point 3 to point 1

(a) Assuming only PdV work, sketch a possible cyclic path for steps I –III on a P-V diagram.

(b) Using the exact differential, state function characteristics of S prove that ∆S for step I is always > 0 for an ideal gas expanding adiabatically and irreversibly from point 1 at TA ,PA to point 2 at TB,PB

(c) Is ∆S for step I > 0 for a non-ideal gas as well? Explain your answer.

3. Problem 4.18 Be sure to state and justify all assumptions made

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10.40 Thermodynamics Fall 2003 Problem Set 3 1. Problem 4.11 – try solving the problem two different ways: a.) using the availability and b.) without using the availability. 2. Problem 14.12 3. a) For the plane with two tangent points in Fig. 5.2b, estimate the value of G, H, A, U,

and S of each phase. Obviously, the grids are course, and difficult to read, so the key is to make all of the values of the thermodynamic properties consistent.

b) Explain the significance of a USV surface, like the one on Figure 5.1, but with a

plane with 4 points of tangency.

4. Problem 5.26. For the properties of liquid water you should refer to the steam table

data provided in: Keenan, Joseph H. et al. “Steam Tables: Thermodynamic Properties of Water Including Vapor, Liquid, and Solid Phases.” Malabar, FL: Krieger Publishing Co.,1992.

Be sure to state and justify all assumptions made

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1. Problem 5.4 parts a, b, c and e. 2. Problem 5.2 3. Problem 5.6

Note: Error in 5th Printing and sooner. Equation describing Newton’s Law in Lagrangian Mechanics should read:

x x

L Lx t x∂ ∂ ∂⎛ ⎞ ⎛ ⎞⎛ ⎞=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

4. Problem 6.2 5. Problem 7.1


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1. Let’s compare the pressure-temperature behavior of two important expansion processes that are used for refrigeration and cooling applications: (i) expansion across a Joule-Thompson valve (ii) expansion in an adiabatic turbine To carryout this comparison it is helpful to construct an equation to express the ratio of how temperature varies with pressure for each process. Defining this ratio as follows:

( )

( )

( / )( / )

process i

process ii

T PT P

α∂ ∂

≡∂ ∂

We will also need to specify the type of fluid to be expanded. To keep things simple, let’s limit our analysis to pure CO2 which can be represented by four constitutive property models whose accuracy will depend on the PVT region where the expansion is carried out. The four models are:

(1) ideal gas (2) The Law of Corresponding States

(Hint: Express α in terms of Z and its derivatives) (3) PR EOS (4) Thermodynamic charts given in Figures 8.12 a,b

(a) What is the initial value of α for each model above for the expansion of pure CO2 from an initial state of 100 bar and 37oC?

(b) Which process would you expect to liquefy a greater fraction of the entering CO2 gas if the final pressure state for the expansion is 10 bar?

(c) What technological and economic issues would be important in selecting a process for practical use?

Physical property data for carbon dioxide are given at the end of the problem set and in Appendix G of the text.

(continued…)

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2. Rocky and Rochelle are having one of their classic arguments regarding thermodynamics and they need your help. Imagine a process where supercritical CO2 gas at 320 K and 100 bar is added to a cylinder while maintaining the total entropy and pressure constant. Rocky says that the temperature will increase and Rochelle says it will decrease. Who is correct? Is it possible that they both could be correct? You may want to start your analysis by developing an expression for the temperature change per mole of component j added to a multicomponent mixture under conditions of constant S, P and Ni [j]. Relevant property data are given at the end of the problem. In formulating your solution, you may want to use the data provided in the Table 4-5 of Tables of Thermal Properties of Gases, Hilsenrath et al., National Bureau of Standards Circular 564,1955. You will have to extrapolate a bit.

3. We would like to make a few modifications to the van der Waals EOS and evaluate how the PVT behavior of a pure fluid might change.

(a) expand the RT/( V-b) term of the vdw EOS as a polynomial power series in b/V

(b) If we truncate the expansion after the (b/V)2 term, how does the compressibility change as a function of density from very low density (ρ = 0) to high density (ρ = 1/b)?

(c) Is it possible for a fluid obeying this modified EOS to reach a Zeno condition where Z = 1? If so, where does this occur?

(d) Assuming that the Zeno condition is met for this pure material, estimate the fugacity coefficient ( φi) at this point. How do you interpret your estimate?

Be sure to state and justify all assumptions made. Image removed due to copyright considerations. Please see “Span, and Wagner. J. Phys. Chem.Ref. Data 25, no. 6 (1996).”

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10.40 Thermodynamics Fall 2003 Problem Set 5 Addendum to Problem Set 5, #2: C.H. Jones, Rocky and Rochelle’s cousin, has contacted Rocky and Rochelle with some new thermodynamic data for CO2 that he came across at the library (given below). You may want to use this new information in formulating your solution to Problem #2 (you will have to extrapolate the data a bit). From Tables of Thermal Properties of Gases, Hilsenrath et al., National Bureau of Standards Circular 564, 1955: Image removed due to copyright considerations. Please see “Hilsenrath, et al. “Tables of ThermalProperties of Gases.” National Bureau of Standards Circular 564 (1955).”

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1. Consider a gas mixture of carbon dioxide and nitrogen at 298 K and 5 bar.

(a) How would you model this mixture in terms of its enthalpy and entropy? Be sure to specify your reference states when describing your model.

(b) Estimate the minimum work required to separate 100 moles of a 40 mole%

nitrogen mixture into pure carbon dioxide and nitrogen.

(c) Would your model need to change if the pressure were increased to 200 bar? If so, how? You are not required to perform detailed calculations, but be sure to give the your approach to solving the problem. Include any parameters that you would need to look up or estimate in order to solve the problem.

2. Given a 3D lattice model, 10x10x10, which is connected to a bath at 298 K and in

which there are 100 non-interacting gas particles having no internal structure, each of which are located only at the lattice points, calculate U, S, G, and Cv.

3. For the same lattice in number 2., but now with only 2 non-interacting particles,

each of which can exist in only two states, such that E1 = 3 kcal/mol and E2 = 5 kcal/mol, calculate U, S, G, and Cv. (Note that this is a simple model of a two-state equilibrium, such as a protein which can exist in either a folded or unfolded state, in solution.)

4. For 0.5 mol of O2 at 0.1 atm., and 1000 K, compute U, S, G, and Cv. You may

treat O2 as an ideal gas at these T and P conditions. Note that both O2 and the O atom have a net spin of 1 in their ground states. Also the first excited electronic state of O2 occurs at a thermal energy of 11,300 K and has a net spin of 1/2. The bond dissociation energy of O2 , Do, is 117.1 kcal/mol and the thermal vibrational and rotational energies are 2230 K and 2.07 K respectively.

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1. Sketch how Cv of xenon, carbon monoxide, and water behave as a function of temperature at low densities. Carefully note the limits as T goes to 0 K and as T gets large (but less than the first electronic excited state).

2. Using MatLab, Excel, or a similar software program, determine the value of N where the error in the Stirling approximation becomes less than 0.1%.

3. Problem 10.1 4. Problems 10.4 and 10.5

5. Problem 10.8 (parts a and b only)


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1. Compute U/(RT) of a mole of diatomic ideal gas molecules treating the vibrational mode classically. Assume that the vibrational frequency is 1013 s-1, and take the reference state as E = 0 when the system is in its ground state (degeneracy of 1 and no important excited electronic states). In a few words, explain the significance of the difference between the quantum and classical results.

2. Show that the standard deviation or square root of the variance of the

distribution of particle densities, σρ, for a pure fluid varies as (<N>)-1/2 for a fixed volume system. For typical thermodynamic systems, what happens to the value of σρ / <ρ> at low densities and at the critical point?

3. Problem 10.11 in the text

4. Problem 10.12 in the text

Be sure to state and justify all assumptions made.

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In this problem set, you will use molecular simulations in order to compute some properties of a simple model of water, TIP3P (CHARMM), as discussed in class. Your calculations will be toy calculations to give you a feel for the application and potential of molecular simulations, without the computational intensity or methodological sophistication of research level computations. 1. For 60 TIP3P (CHARMM) water molecules in a 15Å×15Å×15Å box, compute P,

U, and Cv, at 300 K and at 500K, running for 1000 heat-up steps and 10,000 and 20,000 equilibrium steps. Also, compute the standard deviations of P and U. If you split up the 20,000 step run into 4 pieces and compute CV and the standard deviations for the averages of P and U of the 4 pieces, what do you get? What are the limits of these numbers as the molecular dynamics trajectory lengths get very large? Finally, what is the difference in the meaning of the standard deviations from the whole trajectory and the standard deviations of averages of pieces of the trajectory?

2. Compute at least two P-V isotherms. See if you can find one in the liquid-vapor

coexistence region. If you can find one, estimate the densities of the two phases at coexistence.


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1. Using the Peng-Robinson EOS to calculate the fugacity coefficient of pure carbon dioxide, estimate the vapor pressure and saturated liquid and vapor molar volumes at these temperatures – 10, 20, and 30oC. How do your predictions compare with data for pure CO2 (see Table below)? You might want to refer to Example 8.4 on p. 276-277, Example 9.7 on p. 350-351, and Figure 8.12. If you use Matlab or a similar mathematical software package please document your results and describe the approach you followed in reaching a solution.

Experimental Data for Pure CO2

Temp (oC)

Pressure(bar)

Liquid Volume(cm3/mole)

Vapor Volume (cm3/mole)

10 45.02 51.11 325.6 20 57.29 56.91 226.6 30 72.14 74.18 127.5

2. Problem 15.31 in the textbook. 3. Estimate the critical constants (Tc, Pc), accentric factor (ω) and vapor pressure of

pure caffeine as described in Problem 13.4 of the text. To solve this problem you will have to review the methods presented in Chapter 13 for predicting physical properties using molecular group contributions. (Note: Since the >N– (ring) group is not listed, use the >N– (non-ring) group in its place).

4. Problem 15.38 in the textbook. 5. Problem 15.29 in the textbook.


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10.40 Thermodynamics Fall 2003 Problem Set 11 1. Problem 17.9 in the texbook. Please note that some printings of the book have a typo in Figure P17.9. Specifically, the labels on the univariant loci for the H-H2O(l)-CO2(g) and H-H2O(s)-CO2(g) curves are reversed in some printings. A correct version of the figure is given below.

Image removed due to copyright considerations. Please see “Tester, J. W., and Michael Modell. Thermodynamics and Its Applications. Upper Saddle River, NJ: Prentice HallPTR, 1997, p. 857. Fig. P17.9.” 2. Hydrogen is being considered for widespread deployment as a primary energy carrier. A key motivation for this is to avoid the production and emissions of carbon dioxide associated with the combustion of fossil fuels in power plants and transportation equipment. To achieve this goal, a transition to a non-fossil based primary energy supply system is required. One option is to use nuclear power for supplying all our electricity, heating and transportation fuel needs. In this problem, you are to address the issues raised in parts a-d below in analyzing the effectiveness of three proposed approaches for using nuclear power to produce hydrogen. In all three approaches a new Generation IV fission reactor that uses a high temperature helium gas cooled cycle will be used to provide the energy source in the form of heat and/or electric power. The three approaches are:

1. Direct thermolysis using heat to directly split water into hydrogen and oxygen. 2. High temperature electrolysis using electricity to electrolyze water. 3. Thermochemically using heat in a multi-step cycle.

In answering parts a-d below please clearly state all assumptions made. (a) Referring to approach (1.), helium gas is circulated through the reactor core where it

is heated to 850oC. Using a helium gas stream at this temperature as a means to heat steam, estimate the maximum fraction of steam at 1 bar that will be decomposed by the direct thermolysis reaction:

H2O(g) = H2(g) + 1/2O2(g) (A) Assuming that only reaction (A) occurs, at what temperature would you expect the conversion to hydrogen to be 50% complete? Would increasing the pressure to 200 bar have a measurable effect on conversion at this temperature? If so, describe

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how you would calculate it.

(b) Referring to approach (2.), the addition of electrical work in a properly designed electrolysis cell with Pt electrodes will catalytically decompose water to hydrogen and oxygen as given in reaction A. Estimate the minimum amount of electrical power that would be required to produce 20 moles of hydrogen per second by electrolysis carried out at 850oC and 1 atm.

(c) In a practical electrolysis process, transport and kinetic limitations increase the

required power to 1.5 times the minimum from part (b). Assuming that helium gas at 850oC is fed to the power cycle, estimate the absolute minimum rate of heat transfer from the reactor core needed to supply the required electrical power. An average ambient temperature for heat rejection of 25oC can be assumed.

(d) Referring to approach (3.), the following 3-step thermochemical cycle using iodine (I) and sulfur (S) based intermediates has been proposed by General Atomics:

1. I2 + SO2 + 2H2O = 2 HI + H2SO4 2. 2 HI = H2 + I2 3. H2SO4 = H2O + SO2 + ½ O2

where all components are in a gaseous state. In their process, liquid water is supplied at 25oC and H2 and O2 are produced at 25oC which is also the ambient temperature. General Atomics rates the efficiency of their overall 3-step process at 0.47-- defined as the ratio of the heating value of the net H2 produced to the heat input from the reactor helium loop. To justify whether 0.47 (or 47%) is a thermodynamically acceptable value, estimate the maximum efficiency of the process assuming all reactions in the cycle go to completion. Note that the net cycle given by reactions (1-3) is equivalent to reaction (A) given in part (a).

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Data: Heat capacities of steam (H2O(g)) , He(g), H2(g), and O2(g) can be assumed constant with the following values: Cp (H2O(g) ) = 43.95 /mol K Cp (H2(g)) = 29.3 J/mol K Cp (O2(g)) = 29.3 J/mol K Cp (He(g)) = 20.8J/mol K

Standard Gibbs energy and enthalpy of formation of water: ∆Go

f (H2O(g) ) = -228.8 kJ/mol and ∆Hof (H2O(g) ) = -241.9 kJ/mol with

standard states taken as ideal gas at 1 atm, 298.15 K. Enthalpy of vaporization of pure water at 25oC = ∆Hvap = 2442.5 kJ/kg See also Appendix G and Figures 8.11a and 8.11b in Thermodynamics and its Applications, 3rd ed. for additional property data.

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10.40 Fall 2003 Thermodynamics Exam 1 9 to 11 am Use of Thermodynamics and Its Applications, 3rd ed. and 10.40 class notes and handouts and a hand calculator are permitted. No other aids or documents are permitted. Please answer each question in a separate exam book. Be careful about managing your time and be aware of the maximum points allotted to each part of the 3 problems. 1. (40 points) This is a 6 part problem worth a total of 40 points

(a) (6 points) If for a given system, closed, with one component, we had a function A=A(U,T), could this function contain the equivalent amount of information as the fundamental equation for intensive properties? Briefly explain your answer.

(b) (6 points) Express the following derivative in terms of measurable properties for a

single component: T

GA

∂⎛ ⎞⎜ ⎟∂⎝ ⎠

.

(c) (6 points) Express ( ) ( , , , , ,...)kA B Cy f P S N Nµ= in terms of a Legendre transform

of H = y (0) for an n component system. Also express the total differential, , in terms of derived and primitive thermodynamic properties.

( )kdy

(d) (10 points) Are there restrictions on the signs of Cp and αp for any real, single

component system? Explain your answer. (e) (6 points) For a system containing 5 components, considering only intensive

properties, how many dimensions does the Gibbs surface have, not including the U dimension? Explain the significance of a plane tangent to the Gibbs surface at two points.

(f) (6 points) Express GPT in terms of derivatives of H for an n component system.



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2. (30 points) Technology Review (December 1975) describes a new invention for producing power – the Nitinol Engine. “Nitinol is a unique nickel-titanium alloy whose mechanical and electrical properties change drastically during a crystalline phase change [occurring at] modest temperatures [about 40oC]. The result is a memory for shape: if deformed while cool, nitinol will return to its undeformed shape when warmed. Alternately heating and cooling it causes motion – a conversion of heat to mechanical energy.” The article also contained the illustration shown below and a quote by A.D. Johnson who claims that 30 W of shaft output were generated with a hot water stream flowing at 10 grams/s at 50oC. MITY Industries needs your help to decide whether it should apply for a patent based on Johnson’s invention. Develop an appropriate analysis to evaluate whether the claim is thermodynamically viable. If your analysis shows that the Nitinol engine is thermodynamically viable, is the claimed output possible from a practical point of view? You can assume favorable conditions where a very large flow of cold water is available at 20oC.

Image removed due to copyright considerations. Please see “TechnologyReview 78, no. 2 (December 1975): 19.” 3. (30 points) A French Physicist, Emile Amagat ( 1841-1915) developed a diagram of PV versus P to represent important volumetric behavior for a single component system. Amagat claims that his diagram, illustrated below in reduced coordinates for a general pure substance, provides the same information content as the P versus V diagram we have discussed frequently in class, but that it allows for improved visualization of behavior at low and high pressures or densities. On the Amagat diagram three parabolic curves (C, M, and J) are shown along with a number of isotherms at multiple values of Tc. The outer curve J is the locus of Joule-Thompson coefficient inversion temperatures where the partial derivative of T as a function of P at constant H changes sign.

(a) (10 points) The intermediate curve M is the locus of PV product minima as a function of P. Show that the intercept of this locus with the reduced PV axis at Pr = 0 corresponds to the reduced Boyle temperature of the fluid.



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(b) (10 points) At high pressures the isotherms become nearly parallel. Develop an

analysis based on the VdW EOS that shows this to be reasonable and estimate the limiting value of PV/P as P ∞ .

(c) (10 points) The inner curve C is the locus of points representing liquid and vapor

phase co-existence or the binodal curves meeting at the critical point with a vertical tangent. Is this behavior consistent with the stability criteria described in chapter 7 of the text? Specifically show that the criteria given by Equations (7-16 and 7-39) are satisfied at the critical point. Explain your answer.

16

12

8

4

00 2 4 6 8 10

PV

/PcV

c

P/Pc

Tc2Tc

3Tc4Tc6Tc

M

Amagat diagram: general appearance in reduced coordinate.

C

J

Image by MIT OCW.



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10.40 Fall 2003 Thermodynamics Exam 2 9 to 11 am Use of Thermodynamics and Its Applications, 3rd ed. and 10.40 class notes and handouts and a hand calculator are permitted. No other aids or documents are permitted. Please answer each question in a separate exam book. Be careful about managing your time and be aware of the maximum points allotted to each part of the 6 problems. 1. (20 points) For a 1 mole molecular system that can only occupy any of 4 different

states:

E3 = 11 kcal/mol

E2 = 8 kcal/mol E1 = 8 kcal/mol

Eo= 3 kcal/mol

(a) (4 points) What is U at T = 300 K? (b) (4 points) What is the probability that a given snapshot of the system will have

an energy of 3 kcal/mol at 300 K? (c) (4 points) If each energy state is increased by 2 kcal/mol, what is the

probability that a given snapshot of the system will have an energy of 3 kcal/mol at 300 K?

(d) (4 points) What is U as T gets very large? (e) (4 points) What is U as T gets very small?

2. (10 points) Imagine a lattice consisting of 10000 sites filled with 3000

indistinguishable particles at temperature T. Only one particle can reside at each lattice site and there is no energy of interaction between particles of any type. The initial state is such that the 3000 particles occupy 30% of the lattice volume separated by a partition from the rest of the lattice. Compute the change in G from the initial state to a final equilibrium state.

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3. (15 points) In terms of the grand partition function, Ξ, the PVT equation of state is expressed explicitly in molecular units in Equation (10-57) as lnPV kT= Ξ . Recalling our discussion in class of the Virial equation of state, develop an analysis to determine under what conditions Eq(10-57) would reduce to the ideal gas equation of state for a range of densities. Describe the nature of the intermolecular interactions that exist that will yield this result.

4. (20 points) For a fluid whose intermolecular interactions follow the Sutherland

potential described in Problem Set 8 (Problem 10.12 of the text):

(a) (10 points) Can B(T), the second virial coefficient, be positive or negative or zero? Justify your answer and explain its significance in terms of molecular interactions.

(b) (10 points) Do you expect that this fluid will exhibit a Zeno point condition at

high density (ρ >> 0) where the compressibility factor Z = PV/RT equals unity? Explain the rationale behind your answer.

5. (20 points) A set of different polyatomic, multi-polar molecules all have similar

intermolecular potential functions for their binary interactions, as described approximately by the functional from shown in the figure below.

r2Rσ

+ε

Rσ

−ε

σΦij 0

(a) (10 points) Develop an expression for B(T), the second virial coefficient, for a

particular pure component species.

(b) (10 points) Would you expect all pure fluids in this set of compounds to follow the Law of Corresponding States? What are the appropriate scaling parameters to non-dimensionalize T, P, and V for each fluid that you could use to test your answer with a set of experimental volumetric data? Briefly explain the rationale behind your choice of scaling parameters.

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6. (15 points) The direct evaluation of the partition function for a liquid is difficult to

achieve in practice. Fortunately a range of approximations are available. In the simple cell model for liquids, a unit cell is envisioned to be composed of a single molecule “engaged" inside a rigid lattice formed by the nearest neighbor molecules surrounding the trapped molecule. This entrapped molecule is allowed to wander inside the cell volume (see figure below). Although it is not possible to characterize the interaction energy of the wandering molecule by a single value, the mean field approximation and free volume concepts can be applied. If we bring N molecules together from infinite mutual separations to form a lattice, the system loses potential energy due to attractive forces which can be equated to the “lattice energy” = ΦL. Now we assume that all molecules are at fixed locations in the lattice, except for the wandering molecule which is assumed to preferentially want to locate at the cell center. Thus, on a per molecule basis the attractive part of the potential energy is ΦL/N. In general, ΦL depends on lattice spacing and hence on the volume. Note that while all molecules have the same size described by spheres of diameter σ, the lattice on the other hand has characteristic nearest neighbor distance of a as shown in the figure below.

(a) (8 points) Develop an expression for the configurational integral for the cell in the liquid model.

(b) (7 points) If ΦL/N can be approximated by ΦL = - a/V, where a = a(T) is only a function of temperature, develop a pressure-explicit EOS form in terms of P = f( T, V, Vo , a(T)).

3

x

x x

x

xx

wa a

|a-σ|

Solid Liquid

A two-diamensional representation showing the'wanderer' molecule (W) and 'wall' molecules (X)in a simple cell model.

Vo = N σ3/ 2 V = N a3/ 2

Image by MIT OCW.



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10.40 Thermodynamics Final Exam 3 hr, 9 am – 12 noon Use of Thermodynamics and Its Applications, 3rd ed. and 10.40 class notes and handouts and a hand calculator are permitted. No other aids or documents are permitted. Please answer each question in a separate exam book. Be careful about managing your time and be aware of the maximum points allotted to each part of the 5 problems. 1. (20 points) Rocky and Rochelle Jones are having one of their heated arguments on thermodynamics. This time they are trying to reconcile the theoretical and empirical underpinnings of the so-called K factor charts or nomographs that were used during the last century, before computers were readily available, to characterize equilibrium phase partitioning in multicomponent distillation of hydrocarbons. K is defined as the equilibrium ratio of a particular component yi in the vapor phase to xi in the liquid phase. A typical nomograph is shown below for a set of light alkane and alkene hydrocarbons -- C-1 (methane) through C-9 (nonane), Rocky claims that the K factors do a reasonable job of capturing important non-idealities over a wide range of conditions. In fact, Rocky notes that you only need the temperature and pressure to specify the vapor liquid equilibrium state as indicated by the nomograph. By connecting a straight line to a specific T and P, the intersection of that line with curve for each compound yields the K = yi/xi. Rochelle adamantly maintains that the K’s are only based on single component information and that they fail to include intermolecular interactions. We need your help to settle this argument. (a) (10 points) Derive a general relationship for K in an n-component system vapor-liquid mixture at equilibrium. Clearly indicate the functional dependence of all derived parameters on measurable properties and state what type of constitutive property relationships are needed to evaluate K for each component. (b) (10 points) For the P and T conditions given in the nomograph, who do you think is correct, Rocky or Rochelle? Explain.

(Image removed due to copyright considerations.)

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K-values for systems of light hydrocarbons



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2. (20 points) MITY Industries is trying to decide where to make its next investments and has been approached by Colossal Technology Inc. (CTI), which claims that it has a new approach for generating motive power from hydrogen that is inherently more efficient than the traditional internal combustion (IC) engines that are in use today in our automobiles and trucks. In their scheme, pressurized pure hydrogen at 500 bar, 25oC is electrochemically oxidized with pure oxygen at 1 bar, 25oC to form water in a Fuel Cell Super Electro Converter (FCSEC) that generates electrical power directly. CTI claims that their concept can produce more power than even the most efficient IC engine using the same H2 and O2 feeds because the FCSEC converter is not subject to Carnot limitations. According to CTI, water exits the FCSEC at 1 bar, 25oC, (a) (12 points) What do you think of CTI’s claim? Back up your answer with appropriate thermodynamic analysis and discussion. (b) (8 points) Pressurized methane, available at 500 bar and 25oC, has also been considered as a transportation fuel. CTI claims that they can chemically reform the methane to hydrogen via steam reforming on the vehicle and then use the FCSEC to produce electricity. Would using methane as a fuel introduce any additional thermodynamic limitations over using pure hydrogen? Explain your answer. Steam reforming of methane proceeds by two dominant reaction pathways:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

24 g g g 2 g

2g g 2 g

CH H O CO 3H

CO H O CO H2 g

+ = +

+ = +

Feel free to use the thermochemical and other physical property available in Appendix G and elsewhere in the text.

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3. (20 points) A protein system exists such that each protein molecule is either in its native form (N) or its denatured form (D). The N and D forms have different energy distributions (density of states) with the following structures:

D-form ωD(E) N-form

N D€

(a) (10 points) Explain under what circumstances you would expect the chemical equilibrium constant to follow the van’t Hoff equation (ln K vs. 1/T is a straight line). Back up your explanation with appropriate numerical analysis.

(b) (10 points) Describe briefly using appropriate equations how the situation would differ for the gas phase equilibrium that we discussed in class: H2 + ½ O2 = H2O Recall: the semi-classical canonical partition function is given by NQ

/ ( ) ( )E kTNQ e E dω

∞−

−∞

= ∫ E

where ω(E) is the density of energy states.

ED−σD ED+σD

LD

LN

0ED

EN−σN EN+σN

0

ωN(E)

EN

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4. (20 points) Important reactions between natural minerals occur under high pressure conditions in deep underground reservoirs in the presence of water. For many geologic systems the time scales are so long (typically years to centuries) that equilibrium can be safely assumed. One such system is the gypsum-anhydrite-water equilibrium where

( )

( )( )

( )( )

( )4 2 2s 4 sCaSO 2H O CaSO 2H O

gypsum anhydrite water⋅ = + f

In deriving the criteria for phase equilibrium, we found that the temperature and pressure of all phases were equal. In deeply located mineral systems in rock reservoirs, fluid phase pressures can be different than solid phase pressures if the rock column above the minerals is permeable to fluid. In other words, the hydraulic gradient and lithostatic gradient of pressure can be different. In this case, the system has one more degree of freedom and the equilibrium distribution of products and reactants at any particular depth depends on fluid pressure (Pf) and solid phase pressure (Ps) where both can be expressed as a function of depth z.

P gdz

P gdz

f fo

z

s so

z

=

=

zz

ρ

ρ

(hydraulic)

(lithostatic)

For the gypsum-anhydrite system, let’s consider what happens over a range of conditions. The standard Gibbs free energy change for the reaction is given by Eq. (8-165):

,

1(298K,1bar)

no orx i f i

iG v

=

∆ = ∑ G∆

Compound ,of iG∆

(kJ/mol)water H2O (liquid) -237.141 gypsum CaSO4·2H2O (solid) -1797.20 anhydrite CaSO4 (solid) -1321.70

Reference states for all components are taken to be unit fugacity at 298 K, 1 bar pressure. The molar volume change for the solid phase gypsum to anhydrite transformation is:

→ •∆ = − = −4 4 2

3gypsum anhydrite (CaSO ) (CaSO 2H O) 29.48cm /molV V V

and

2

3H O, liquid=18.07 cm molV

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Over accessible depths, pressures ranged from 1 bar to as high as 1000 bar while temperatures can range from near ambient to 200°C. (a) (10 points) Assuming that the standard state is now redefined at the temperature and pressure of the system, how will ∆Grx

o be affected by pressure at 298 K? Express your answer in terms of ∆Grx

o = f(Pf, Ps). (b) (10 points) For the cases of ρs/ρf = 3.0 and ρs = ρf ; describe how you could estimate the variation of Pf with T for the gypsum to anhydrite transformation. What other property information is required to make the estimate? In both cases ρf can be taken as 1000 kg/m3.

[ Hint: consider using the reaction equilibrium constraint Σ viµi = 0 directly to determine a relationship between Pf and T assuming that all phases remain pure.]

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5. (20 points) As described in class, we saw that the osmotic virial expansion of McMillan and Mayer provides a way to determine the osmotic pressure (Π) of a binary aqueous solutions for a range of solutes. Here we find

( )2 32 2 2 3 2

22

2

...

A

B BkT

N CM

ρ ρ ρ

ρ

Π= + + +

≡

11-48

where NA = Avogadro’s number , C2 = solute concentration in g/L, M2 = molecular weight of the solute, B2 = second osmotic virial coefficient, B3 = third osmotic virial coefficient, etc. We can express the activity of the water solvent as ln aw at some temperature, T, a reference pressure of 1 atm, and some concentration of the solvent:

ln ww

VaRT

Π= −

where wV = molar volume of solvent (water). Cryobiology researchers (J. Elliott, et al, 2003) have recently discovered that the osmotic virial equation can also be used to accurately estimate freezing point depression for a range of aqueous based biological systems. For example, freezing point depressions are correctly determined for macromolecular species such as hemoglobin and bovine serum albumin. The figure below shows experimental freezing point depression data for hemoglobin in water. (a) (10 points) Develop an expression for the freezing point depression using the osmotic virial expansion. (Hint – both freezing point depression and osmotic pressure are colligative effects where the presence of solute lowers the chemical potential of the solvent in a similar manner) (b) (10 points) By making appropriate assumptions regarding solute-solute interactions, use the freezing point depression data provided to estimate the excluded volume of a hemoglobin molecule. Be sure to clearly state the assumptions that you make.

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Data:

−

−

= ×

= ×

∆ =

=

= =

3

3

,

0

1.000 10 L/g

1.091 10 L/g335 J/g

64,000 g/mol

T freezing point of pure water at 1 atm 273.15 K

LwIce

w

fus w

Hemoglobin

V

VH

M

Freezing Point Depression for Hemoglobin

-2.0E-04

-1.8E-04

-1.6E-04

-1.4E-04

-1.2E-04

-1.0E-04

-8.0E-05

-6.0E-05

-4.0E-05

-2.0E-05

0.0E+00

2.0E-050 200 400 600 800

C2 (g/L)

(T-T

0)/C

2 (K

*L/g

)

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